Question

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Write the standard form of the quadratic function whose graph is a parabola with the given vertex and that passes through the given point. (Let x be the independent variable and y be the dependent variable.)

Vertex: (1, -5); point: (-1, 7)

Answer 1

y = 3(x - 1)^2 - 5

Explanation:

Vertex formula: y=a(x+h)^2+k where h is the change in x,

k is the change in y, and a is the coefficient of the function

k=-5

x+h=0

1+h=0 ==> plugin x=1

h=-1

Hence, the equation is:

y=a(x+(-1))^2+(-5)

y=a(x-1)^2 - 5 ==> solve for a

7=a((-1)-1)^2 - 5 ==> plugin point (-1, 7) where x=-1 and y=7

7=a(-2)^2 - 5 ==> simplify

7=4a - 5 ==> (-2)^2 = 2^2 = 4

12 = 4a ==> add 5 on both sides to isolate a

a = 3 ==> divide 4 on both sides to isolate a

Answer: y=3(x-1)^2 - 5

Answer 2

• f(x) = 3x² - 6x - 2

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Standard form of quadratic function:

• f(x) = ax² + bx + c, where a, b, c - constants

Vertex form of quadratic function:

• f(x) = a(x - h)² + k, where (h, k) is coordinates of the vertex, a- constant

Given vertex (1, - 5), substitute into vertex form:

• f(x) = a( x - 1)² - 5

Given the point (-1, 7) on the graph, substitute into equation and find the value of a:

• 7 = a( - 1 - 1)² - 5
• 7 = 4a - 5
• 4a = 12
• a = 3

Now, substitute the value of a and convert the equation into standard form:

• f(x) = 3(x - 1)² - 5
• = 3(x² -2x + 1) - 5
• = 3x² - 6x + 3 - 5
• = 3x² - 6x - 2