quadratic function in standard form

Question

Write a quadratic function in standard form whose graph has a vertex of (2.5, 5)​ and passes through (0, 0)​ and (5, 0)​ 

f(x)=

Answer 1

You are given both x intercepts (0,0), (5,0) and the vertex (2.5, 5).

Steps Reasoning

Factors = (x-0) and (x-5) ..........1. Use the x intercepts (0,0), (5, 0) to find the factors

y = a(x-0)(x-5) ..........................2. Write the factored form of of the quadratic. Include the coefficient a. This is just a number.

y = ax(x-5) ................................3. replace (x-0) with x

y = a(x² - 5x) ............................4. distribute the x to get the equation in quadratic form.

5 = a (2.5² - 5 ⋅ 2.5) .................5. replace x and y with the (x, y) coordinates (2.5, 5)

5 = a (6.25 - 12.5) ....................6. Solve for a

5 = a (-6.25)

5 / -6.25 = a

a = -0.8

y = -0.8 (x² - 5x) ........................7. Write the equation with a

y = -0.8 x² + 4x ..........................8. The equation in standard form.

Standard form is y = ax² + bx + c

In this case a = -0.8, b = 4 and c = 0.

Answer 2

Vertex Form of a Parabola

y=a(x-h)²+k with vertex (h,k) and growth/shrink factor "a"

Solve for "a" using one of the given points

y=a(x-2.5)²+5

0=a(0-2.5)²+5 <-- (x,y)=(0,0)

0=6.25a+5

-5=6.25a

-0.8=a

In vertex form, this would be y=-0.8(x-2.5)²+5

In standard form, this would be y=-0.8x²+4x

Hope this helped!

Answer 3

f(x) = a(x-2.5)^2+5

(0,0) on graph implies 0 = a(0-2.5)^2+5

0=6.25a+5

-5=6,25a implies a = -0.8

f(x) = -0.8(x-2.5)^2+5

The point (0,0) is on the graph but provides no additional information.