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

Explanation:

Okay, let me walk through solving both of these problems to find the standard form of the quadratic equations:

Problem #1:

Given:

Vertex: (-2,6)

Point: (-4,-2)

Standard form: y = a(x - h)2 + k

Where (h, k) is the vertex

Plug in the vertex:

h = -2

k = 6

Plug in the point:

At x=-4, y=-2

-2 = a((-4)-(-2))2 + 6

-2 = a(-2)2 + 6

-2 = 4a + 6

Solve for a:

-8 = 4a

a = 2

Therefore, the standard form is:

y = 2(x + 2)2 + 6

Problem #2:

Given:

Y-intercept: (0,3)

Vertex: (x,-1)

Point: (2,3)

Y-intercept gives us k=3

Vertex x value gives us h=x

Vertex y value gives us k=-1

Plug into standard form:

y = a(x - x)2 - 1

Plug in point:

At (2,3):

3 = a(2 - x)2 - 1

3 = a(2)2 - 1

3 = 4a - 1

4a = 4

a = 1

The standard form is:

y = 1(x - x)2 - 1