Question

I need the standard form on both of these problems: Problem #1: find the standard form given the vertex (-2,6) and a point on the graph being (-4,-2). Problem #2: find the standard form given the y intercept at (0,3) and the vertex at (x,-1) and a point at (2,3)

Answer 1

The standard form of both equations are y = -2x² - 8x - 2 and y = x² -2x + 3

How to determine the standard form of both equations

From the question, we have the following parameters that can be used in our computation:

Problem 1

Vertex = (-2,6)

Point = (-4,-2).

A quadratic function is represented as

y = a(x - h)² + k

Where

(h, k) = vertex

So, we have

y = a(x + 2)² + 6

Using the points, we have

a(-4 + 2)² + 6 = -2

4a = -8

a = -2

So, we have

y = -2(x + 2)² + 6

Expand

y = -2(x² + 4x + 4) + 6

y = -2x² - 8x - 8 + 6

y = -2x² - 8x - 2

Problem 2

y-intercept = (0, 3)

Point = (0, 3) and (2, 3)

A quadratic function is represented as

y = ax² +bx + c

Where

c = y-intercept

i.e c = 3

So, we have

y = ax² +bx + 3

Using the points, we have

a(2)² + 2b + 3 = 3

4a + 2b = 0

2a + b = 0

b = -2a

Set x = 1

So, we have

b = -2

Recall that

y = ax² +bx + 3

This gives

y = x² -2x + 3

Hence, the standard form of both equations are y = -2x² - 8x - 2 and y = x² -2x + 3