💡Okay, so I need to figure out the sum of the measures of the six interior angles of a hexagon. Hmm, I remember that polygons have different properties, and there's a formula to find the sum of their interior angles. Let me try to recall that formula.I think it's something like (n-2) times 180 degrees, where n is the number of sides. Yeah, that sounds right. So, for a triangle, which has 3 sides, the sum would be (3-2)*180 = 180 degrees, which makes sense because a triangle's angles add up to 180. For a quadrilateral, like a square, it would be (4-2)*180 = 360 degrees, which also checks out because a square has four right angles, each 90 degrees, and 4*90 is 360.Okay, so applying this formula to a hexagon, which has 6 sides. Let me plug in the numbers. So, n is 6, so (6-2)*180. That would be 4*180. Let me calculate that: 4 times 180 is 720. So, the sum of the interior angles of a hexagon should be 720 degrees.Wait, let me double-check to make sure I didn't make a mistake. Maybe I can visualize a hexagon. A regular hexagon has all sides and angles equal, right? Each interior angle in a regular hexagon is 120 degrees because 720 divided by 6 is 120. Yeah, that makes sense. So, if each angle is 120 degrees, six of them would add up to 720 degrees.Is there another way to think about this? Maybe by dividing the hexagon into triangles. If I draw lines from one vertex to all the other non-adjacent vertices, I can split the hexagon into triangles. For a hexagon, that would create 4 triangles, right? Because n-2, which is 6-2, equals 4. Each triangle has 180 degrees, so 4 triangles would have 4*180 = 720 degrees. That matches what I got earlier.I guess another way to confirm is to think about the exterior angles. I remember that the sum of the exterior angles of any polygon is always 360 degrees. But wait, that's for the exterior angles. The interior and exterior angles at each vertex add up to 180 degrees because they are supplementary. So, if I know the sum of the exterior angles is 360, then the sum of the interior angles would be (n*180) - 360. For a hexagon, that's (6*180) - 360 = 1080 - 360 = 720 degrees. Yep, that still gives me the same answer.I think I'm confident now that the sum of the interior angles of a hexagon is 720 degrees. It matches with the formula, the division into triangles, and even the relationship with exterior angles. So, I don't think I made any mistakes here.

💡First, I need to determine the first integer in the sequence of 10 consecutive integers. Let's denote this first integer as ( x ).Since the integers are consecutive, the sequence will be ( x, x+1, x+2, ldots, x+9 ).The average of these 10 integers is given as 25, so their total sum is ( 25 times 10 = 250 ).The sum of the consecutive integers can be expressed as:[ x + (x+1) + (x+2) + ldots + (x+9) = 250 ]Simplifying this, we get:[ 10x + (1+2+3+ldots+9) = 250 ]The sum of the first 9 positive integers is ( frac{9 times 10}{2} = 45 ), so:[ 10x + 45 = 250 ]Solving for ( x ):[ 10x = 250 - 45 ][ 10x = 205 ][ x = 20.5 ]Next, we deduct 9 from the first number, 8 from the second, and so on, until 0 from the last number. The new sum of the numbers will be:[ (x-9) + (x+1-8) + (x+2-7) + ldots + (x+9-0) ]This simplifies to:[ (x-9) + (x-7) + (x-5) + ldots + (x+9) ]The deductions form an arithmetic series with a common difference of 1, starting from 9 and ending at 0, which sums up to:[ 9 + 8 + 7 + ldots + 0 = frac{9 times (9+1)}{2} = frac{90}{2} = 45 ]The new sum of the numbers after the deductions is:[ 250 - 45 = 205 ]Finally, the new average is the new sum divided by the number of integers, which is 10:[ text{New average} = frac{205}{10} = 20.5 ]

💡Okay, so I have this geometry problem here about a semicircle with diameter AB, where AB is 100 units long. Points C and D are on the semicircle, and I know the lengths AC is 28 and BD is 60. I need to find the length of CD. Hmm, let me try to visualize this.First, I should probably draw a diagram. So, there's a semicircle with AB as the diameter. Let me mark points A and B at the ends of the diameter. The center of the semicircle would be the midpoint of AB, which I'll call O. Since AB is 100, the radius of the semicircle is 50, so OA and OB are both 50 units each.Now, points C and D are somewhere on the semicircle. AC is 28, so point C is somewhere on the semicircle such that the straight-line distance from A to C is 28. Similarly, BD is 60, so point D is on the semicircle such that the straight-line distance from B to D is 60. I need to find CD, the distance between points C and D.I think I can use coordinate geometry here. Let me place the semicircle on a coordinate system with the center O at the origin (0,0). Then, point A would be at (-50, 0) and point B at (50, 0). The semicircle would be the upper half of the circle with radius 50.Now, let me denote the coordinates of point C as (x1, y1) and point D as (x2, y2). Since both C and D lie on the semicircle, they must satisfy the equation of the circle: x² + y² = 50², which is 2500. Also, since it's a semicircle, y is positive for both points.Given that AC is 28, the distance between A (-50, 0) and C (x1, y1) is 28. So, using the distance formula:√[(x1 + 50)² + (y1 - 0)²] = 28Squaring both sides:(x1 + 50)² + y1² = 784But since x1² + y1² = 2500, I can substitute that in:(x1 + 50)² + (2500 - x1²) = 784Expanding (x1 + 50)²:x1² + 100x1 + 2500 + 2500 - x1² = 784Simplify:100x1 + 5000 = 784100x1 = 784 - 5000 = -4216x1 = -4216 / 100 = -42.16Hmm, so x1 is -42.16. Let me find y1 using the circle equation:y1² = 2500 - x1² = 2500 - (-42.16)²Calculating (-42.16)²:42.16 * 42.16. Let me compute that:42 * 42 = 176442 * 0.16 = 6.720.16 * 42 = 6.720.16 * 0.16 = 0.0256So, (42 + 0.16)² = 42² + 2*42*0.16 + 0.16² = 1764 + 13.44 + 0.0256 = 1777.4656So, y1² = 2500 - 1777.4656 = 722.5344Therefore, y1 = √722.5344 ≈ 26.88So, point C is approximately (-42.16, 26.88)Now, let's find point D. BD is 60, so the distance between B (50, 0) and D (x2, y2) is 60.Using the distance formula:√[(x2 - 50)² + (y2 - 0)²] = 60Squaring both sides:(x2 - 50)² + y2² = 3600Again, since x2² + y2² = 2500, substitute:(x2 - 50)² + (2500 - x2²) = 3600Expanding (x2 - 50)²:x2² - 100x2 + 2500 + 2500 - x2² = 3600Simplify:-100x2 + 5000 = 3600-100x2 = 3600 - 5000 = -1400x2 = (-1400)/(-100) = 14So, x2 is 14. Now, find y2:y2² = 2500 - x2² = 2500 - 14² = 2500 - 196 = 2304Therefore, y2 = √2304 = 48So, point D is at (14, 48)Now, I need to find the distance between points C (-42.16, 26.88) and D (14, 48). Using the distance formula:CD = √[(14 - (-42.16))² + (48 - 26.88)²]Simplify the differences:14 - (-42.16) = 14 + 42.16 = 56.1648 - 26.88 = 21.12So, CD = √[(56.16)² + (21.12)²]Calculating each square:56.16²: Let's compute 56² = 3136, 0.16² = 0.0256, and cross term 2*56*0.16 = 17.92So, (56 + 0.16)² = 56² + 2*56*0.16 + 0.16² = 3136 + 17.92 + 0.0256 ≈ 3153.9456Similarly, 21.12²: 21² = 441, 0.12² = 0.0144, cross term 2*21*0.12 = 5.04So, (21 + 0.12)² = 21² + 2*21*0.12 + 0.12² = 441 + 5.04 + 0.0144 ≈ 446.0544Adding these together:3153.9456 + 446.0544 = 3600So, CD = √3600 = 60Wait, that's interesting. So, CD is 60. But BD is also 60. Hmm, so CD equals BD? That seems a bit coincidental, but maybe it's correct.Let me double-check my calculations to make sure I didn't make any errors.First, for point C:AC = 28, so distance from A (-50,0) to C (x1,y1):√[(x1 + 50)² + y1²] = 28Squared: (x1 + 50)² + y1² = 784But since x1² + y1² = 2500, subtracting gives:(x1 + 50)² - x1² = 784 - 2500Which is:x1² + 100x1 + 2500 - x1² = -1716So, 100x1 + 2500 = -1716100x1 = -1716 - 2500 = -4216x1 = -42.16, which is correct.Then, y1² = 2500 - (-42.16)² ≈ 2500 - 1777.4656 ≈ 722.5344, so y1 ≈ 26.88, correct.For point D:BD = 60, so distance from B (50,0) to D (x2,y2):√[(x2 - 50)² + y2²] = 60Squared: (x2 - 50)² + y2² = 3600Since x2² + y2² = 2500, subtracting gives:(x2 - 50)² - x2² = 3600 - 2500 = 1100Expanding (x2 - 50)²:x2² - 100x2 + 2500 - x2² = -100x2 + 2500 = 1100So, -100x2 = 1100 - 2500 = -1400x2 = 14, correct.Then, y2² = 2500 - 14² = 2500 - 196 = 2304, so y2 = 48, correct.Now, distance CD:C (-42.16, 26.88) to D (14,48):Δx = 14 - (-42.16) = 56.16Δy = 48 - 26.88 = 21.12Distance squared: (56.16)² + (21.12)²As calculated earlier, this is approximately 3153.9456 + 446.0544 = 3600So, distance CD = √3600 = 60Therefore, CD is indeed 60.But wait, BD is also 60. So, does that mean CD equals BD? That seems to be the case here.Is there a geometric reason why CD equals BD? Let me think.Looking back, point D is such that BD = 60. If CD also equals 60, then triangle CBD is isoceles with CB = CD. But I'm not sure if that's necessarily the case.Alternatively, maybe points C and D are arranged in such a way that CD equals BD. But in this specific case, with the given distances AC = 28 and BD = 60, it turned out that CD is 60.I think the calculations are correct, so the answer should be 60.**Final Answer**boxed{60}

💡Okay, so I need to prove that there exists a rational coefficient polynomial function ( y = x^3 + b x^2 + c x + d ) which has three distinct rational roots, and whose critical points are also rational. Hmm, let me break this down step by step.First, I know that a cubic polynomial with rational coefficients can have rational roots if it factors into linear terms with rational roots. So, if I can find such a polynomial, that would satisfy the first condition. But I also need the critical points to be rational. Critical points are where the derivative is zero, so I need the derivative to have rational roots as well.Let me recall that the derivative of a cubic polynomial ( y = x^3 + b x^2 + c x + d ) is ( y' = 3x^2 + 2b x + c ). For the critical points to be rational, the roots of this quadratic equation must be rational. That means the discriminant of the quadratic must be a perfect square. The discriminant is ( (2b)^2 - 4 cdot 3 cdot c = 4b^2 - 12c ). So, ( 4b^2 - 12c ) must be a perfect square.Now, I need to construct such a polynomial. Maybe I can start by choosing some rational roots and then see if the critical points are also rational. Let me pick three distinct rational numbers as roots. For simplicity, let's choose -3, 0, and 5. So, the polynomial would be ( y = (x + 3)(x)(x - 5) ). Let me expand this:First, multiply ( (x + 3)(x) ):( x(x + 3) = x^2 + 3x ).Now, multiply that by ( (x - 5) ):( (x^2 + 3x)(x - 5) = x^3 - 5x^2 + 3x^2 - 15x = x^3 - 2x^2 - 15x ).So, the polynomial is ( y = x^3 - 2x^2 - 15x ). Let me check its critical points by taking the derivative:( y' = 3x^2 - 4x - 15 ).To find the critical points, set ( y' = 0 ):( 3x^2 - 4x - 15 = 0 ).Using the quadratic formula:( x = frac{4 pm sqrt{(-4)^2 - 4 cdot 3 cdot (-15)}}{2 cdot 3} = frac{4 pm sqrt{16 + 180}}{6} = frac{4 pm sqrt{196}}{6} = frac{4 pm 14}{6} ).So, the solutions are:( x = frac{4 + 14}{6} = frac{18}{6} = 3 ) and ( x = frac{4 - 14}{6} = frac{-10}{6} = -frac{5}{3} ).Both critical points are rational numbers. Therefore, the polynomial ( y = x^3 - 2x^2 - 15x ) satisfies the conditions: it has three distinct rational roots (-3, 0, 5) and its critical points at ( x = 3 ) and ( x = -frac{5}{3} ) are also rational.Wait, but is this the only way? Maybe I can think of another polynomial. Let me try with different roots. Suppose I choose 1, 2, and 3 as roots. Then the polynomial would be ( y = (x - 1)(x - 2)(x - 3) ). Let me expand this:First, multiply ( (x - 1)(x - 2) ):( x^2 - 3x + 2 ).Now, multiply by ( (x - 3) ):( (x^2 - 3x + 2)(x - 3) = x^3 - 3x^2 - 3x^2 + 9x + 2x - 6 = x^3 - 6x^2 + 11x - 6 ).Taking the derivative:( y' = 3x^2 - 12x + 11 ).Set ( y' = 0 ):( 3x^2 - 12x + 11 = 0 ).Using the quadratic formula:( x = frac{12 pm sqrt{144 - 132}}{6} = frac{12 pm sqrt{12}}{6} = frac{12 pm 2sqrt{3}}{6} = 2 pm frac{sqrt{3}}{3} ).Hmm, these are not rational numbers. So, the critical points are irrational in this case. That means not all polynomials with rational roots will have rational critical points. So, I need to be careful in choosing the roots.Maybe there's a pattern or a specific condition that the roots must satisfy for the critical points to be rational. Let me think about the relationship between the roots and the critical points.Given a cubic polynomial ( y = x^3 + b x^2 + c x + d ), the critical points are the solutions to ( 3x^2 + 2b x + c = 0 ). The roots of this quadratic are given by:( x = frac{-2b pm sqrt{4b^2 - 12c}}{6} = frac{-b pm sqrt{b^2 - 3c}}{3} ).For these to be rational, ( sqrt{b^2 - 3c} ) must be rational. Since ( b ) and ( c ) are rational, ( b^2 - 3c ) must be a perfect square of a rational number.Alternatively, if I express the polynomial in terms of its roots, ( y = (x - alpha)(x - beta)(x - gamma) ), then expanding gives:( y = x^3 - (alpha + beta + gamma)x^2 + (alphabeta + betagamma + gammaalpha)x - alphabetagamma ).So, ( b = -(alpha + beta + gamma) ), ( c = alphabeta + betagamma + gammaalpha ), and ( d = -alphabetagamma ).The derivative is:( y' = 3x^2 - 2(alpha + beta + gamma)x + (alphabeta + betagamma + gammaalpha) ).Setting this equal to zero:( 3x^2 - 2(alpha + beta + gamma)x + (alphabeta + betagamma + gammaalpha) = 0 ).The discriminant is:( [2(alpha + beta + gamma)]^2 - 4 cdot 3 cdot (alphabeta + betagamma + gammaalpha) = 4(alpha + beta + gamma)^2 - 12(alphabeta + betagamma + gammaalpha) ).Simplifying:( 4[(alpha + beta + gamma)^2 - 3(alphabeta + betagamma + gammaalpha)] ).This discriminant must be a perfect square for the critical points to be rational. Let me compute this expression:( (alpha + beta + gamma)^2 - 3(alphabeta + betagamma + gammaalpha) = alpha^2 + beta^2 + gamma^2 + 2alphabeta + 2betagamma + 2gammaalpha - 3alphabeta - 3betagamma - 3gammaalpha = alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha ).So, the discriminant is ( 4(alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha) ). For this to be a perfect square, ( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha ) must be a perfect square divided by 4.Alternatively, if I can choose ( alpha, beta, gamma ) such that ( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha ) is a perfect square, then the discriminant will be a perfect square, ensuring rational critical points.Let me try choosing ( alpha = 0 ), ( beta = 1 ), and ( gamma = -1 ). Then:( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha = 0 + 1 + 1 - 0 - (-1) - 0 = 2 + 1 = 3 ). Hmm, 3 is not a perfect square.Wait, maybe I need to choose roots such that this expression is a perfect square. Let me try ( alpha = 1 ), ( beta = 2 ), ( gamma = 3 ):( 1 + 4 + 9 - 2 - 6 - 3 = 14 - 11 = 3 ). Again, 3 is not a perfect square.How about ( alpha = 1 ), ( beta = 1 ), ( gamma = -2 ):( 1 + 1 + 4 - 1 - (-2) - (-2) = 6 - 1 + 2 + 2 = 9 ). Oh, 9 is a perfect square. So, this might work.Let me construct the polynomial with roots 1, 1, and -2. Wait, but the roots need to be distinct. So, I can't have two roots as 1. Maybe I need to adjust.Let me try ( alpha = 2 ), ( beta = 3 ), ( gamma = -5 ):( 4 + 9 + 25 - 6 - (-15) - (-10) = 38 - 6 + 15 + 10 = 57 ). Not a perfect square.Hmm, this is tricky. Maybe I should look for symmetric roots or something. Let me try ( alpha = a ), ( beta = b ), ( gamma = -(a + b) ). Then:( alpha + beta + gamma = 0 ), so ( b = 0 ). Wait, no, ( gamma = -(a + b) ), so ( alpha + beta + gamma = 0 ). Then, the expression becomes:( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha ).Substituting ( gamma = -(a + b) ):( a^2 + b^2 + (a + b)^2 - ab - b(-a - b) - (-a - b)a ).Let me compute this:( a^2 + b^2 + a^2 + 2ab + b^2 - ab + ab + b^2 + a^2 + ab ).Wait, this seems messy. Maybe it's better to compute it step by step.First, ( gamma = -(a + b) ).Compute ( alpha^2 + beta^2 + gamma^2 ):( a^2 + b^2 + (a + b)^2 = a^2 + b^2 + a^2 + 2ab + b^2 = 2a^2 + 2b^2 + 2ab ).Compute ( alphabeta + betagamma + gammaalpha ):( ab + b(-a - b) + (-a - b)a = ab - ab - b^2 - a^2 - ab ).Wait, let me compute each term:- ( alphabeta = ab )- ( betagamma = b(-a - b) = -ab - b^2 )- ( gammaalpha = (-a - b)a = -a^2 - ab )So, summing these:( ab - ab - b^2 - a^2 - ab = -a^2 - b^2 - ab ).Therefore, the expression ( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha ) becomes:( (2a^2 + 2b^2 + 2ab) - (-a^2 - b^2 - ab) = 2a^2 + 2b^2 + 2ab + a^2 + b^2 + ab = 3a^2 + 3b^2 + 3ab ).So, the discriminant is ( 4(3a^2 + 3b^2 + 3ab) = 12(a^2 + b^2 + ab) ).For this to be a perfect square, ( a^2 + b^2 + ab ) must be a perfect square divided by 12. Hmm, this seems complicated. Maybe I can choose specific values for ( a ) and ( b ) such that ( a^2 + b^2 + ab ) is a perfect square.Let me try ( a = 1 ), ( b = 1 ):( 1 + 1 + 1 = 3 ). Not a perfect square.( a = 1 ), ( b = 2 ):( 1 + 4 + 2 = 7 ). Not a perfect square.( a = 2 ), ( b = 3 ):( 4 + 9 + 6 = 19 ). Not a perfect square.( a = 3 ), ( b = 6 ):( 9 + 36 + 18 = 63 ). Not a perfect square.Hmm, maybe this approach isn't working. Let me think differently.I remember that in the first example I tried, with roots -3, 0, 5, the critical points were rational. So, maybe I can generalize that.In that case, the roots were -3, 0, and 5. Let me compute ( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha ):( (-3)^2 + 0^2 + 5^2 - (-3)(0) - 0(5) - 5(-3) = 9 + 0 + 25 - 0 - 0 + 15 = 49 ). Oh, 49 is a perfect square. So, that's why the discriminant was a perfect square, leading to rational critical points.So, in this case, ( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha = 49 ), which is 7 squared. Therefore, the discriminant was ( 4 times 49 = 196 ), which is 14 squared.So, the key was choosing roots such that ( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha ) is a perfect square.Therefore, to construct such a polynomial, I can choose three distinct rational numbers ( alpha, beta, gamma ) such that ( alpha^2 + beta^2 + gamma^2 - alphabeta - betagamma - gammaalpha ) is a perfect square.Alternatively, since this expression is equal to ( frac{1}{2}[(alpha - beta)^2 + (beta - gamma)^2 + (gamma - alpha)^2] ), which is always non-negative, I can ensure it's a perfect square by choosing the roots appropriately.So, in summary, to prove the existence, I can provide an example of such a polynomial, as I did with roots -3, 0, and 5, which results in a polynomial with rational coefficients, three distinct rational roots, and rational critical points.Alternatively, I can argue that such polynomials exist by showing that there are choices of rational roots where the discriminant of the derivative is a perfect square, ensuring rational critical points.Therefore, the existence is proven by construction.

💡Okay, so I have this problem about an isosceles right-angled triangle with legs of length 2. A point is chosen inside this triangle, and I need to find the probability that the distance from this point to the right-angled vertex is not greater than 1. The options are given as fractions of pi, which makes me think that the solution involves some area calculations, probably involving a circle.First, let me visualize the triangle. Since it's an isosceles right-angled triangle, the two legs are equal, each of length 2. The right angle is at the vertex where these two legs meet. Let me denote this vertex as point A. The other two vertices, B and C, are at the ends of the legs.Now, the problem is about choosing a random point inside this triangle such that the distance from this point to vertex A is less than or equal to 1. So, essentially, I need to find the area within the triangle where all points are within a distance of 1 from point A, and then divide that by the total area of the triangle to get the probability.Let me start by calculating the area of the triangle. Since it's a right-angled triangle, the area is (base * height)/2. Here, both the base and height are 2, so the area is (2*2)/2 = 2. That seems straightforward.Next, I need to figure out the region within the triangle where the distance from point A is less than or equal to 1. This region is a quarter-circle with radius 1, because in a right-angled triangle, the set of points within a certain distance from the right angle vertex forms a quarter-circle. The radius of this circle is 1, so the area of a full circle would be π*(1)^2 = π. But since it's a quarter-circle, the area is π/4.Wait, hold on. Is the entire quarter-circle inside the triangle? Because the legs are of length 2, and the radius is 1, which is less than 2, so yes, the quarter-circle should fit entirely within the triangle. That means the area where the distance from A is less than or equal to 1 is indeed π/4.Now, to find the probability, I need to divide the area of this quarter-circle by the area of the entire triangle. So, probability P = (π/4) / 2 = π/8.Hmm, let me double-check that. The area of the quarter-circle is π/4, and the area of the triangle is 2. So, π/4 divided by 2 is π/8. That seems correct.Wait, but is the quarter-circle entirely within the triangle? Let me think. The triangle has legs of length 2, so from point A, moving 1 unit along both legs, the quarter-circle would extend from A to 1 unit along each leg, and the arc would curve towards the hypotenuse. Since 1 is less than 2, it doesn't reach the hypotenuse, so yes, the quarter-circle is entirely within the triangle.Therefore, the area of the region where the distance from A is ≤1 is π/4, and the total area is 2, so the probability is π/8.Looking back at the options, B is π/8, so that should be the correct answer.**Final Answer**boxed{B}

💡Okay, so I need to find the equation of a parabola. The problem says it has a vertex at (3, -2), a vertical axis of symmetry, and it passes through the point (4,1). I have to express the answer in the form "ax² + bx + c". Hmm, let's see.First, I remember that the vertex form of a parabola is useful when we know the vertex. Since the axis of symmetry is vertical, the parabola opens either up or down. The vertex form is usually written as y = a(x - h)² + k, where (h, k) is the vertex. So in this case, h is 3 and k is -2. That means the equation starts as y = a(x - 3)² - 2.Now, I need to find the value of 'a'. To do that, I can use the point (4,1) that lies on the parabola. If I plug x = 4 and y = 1 into the equation, I should be able to solve for 'a'.Let's substitute those values in:1 = a(4 - 3)² - 2Simplify inside the parentheses first:1 = a(1)² - 2Then square the 1:1 = a(1) - 2So that simplifies to:1 = a - 2To solve for 'a', I can add 2 to both sides:1 + 2 = aWhich gives:3 = aOkay, so 'a' is 3. Now, plugging this back into the vertex form, the equation becomes:y = 3(x - 3)² - 2But the problem asks for the equation in standard form, which is ax² + bx + c. So I need to expand this vertex form into standard form.Let's start by expanding (x - 3)²:(x - 3)² = x² - 6x + 9Now, multiply this by 3:3(x² - 6x + 9) = 3x² - 18x + 27Then, subtract 2:3x² - 18x + 27 - 2 = 3x² - 18x + 25So, putting it all together, the equation in standard form is:y = 3x² - 18x + 25Let me double-check my work to make sure I didn't make any mistakes. Starting from the vertex form, plugging in the point (4,1) gave me 'a' as 3, which seems correct. Then expanding (x - 3)² correctly gives x² - 6x + 9. Multiplying by 3 gives 3x² - 18x + 27, and subtracting 2 results in 3x² - 18x + 25. That looks right.I can also verify by plugging the point (4,1) back into the standard form equation to see if it holds true.Calculating y when x = 4:y = 3(4)² - 18(4) + 25y = 3(16) - 72 + 25y = 48 - 72 + 25y = (48 + 25) - 72y = 73 - 72y = 1Yes, that works. So the equation is correct.I think I've got it. The standard form of the parabola with the given vertex and passing through the given point is y = 3x² - 18x + 25.