How do you find the vertex and intercepts for y = -2(x+1)^2 +7 ?

1 Answer
Jan 14, 2016

Explanation is given below.

Explanation:

Added by Tony BAdded by Tony B

The given problem is already in the vertex form.

color(blue)"The vertex form"

color(maroon)(y=a(x-h)^2+k)

Where (h,k) is the vertex.

Our problem
y=-2(x+1)^2+7
y=-2(x-(-1))^2+7

(h,k) = (-1,7)

The vertex is (-1,7)

Intercepts on x and y axes occur where the curve crosses them.

To find y intercept we need to plug in x=0

y=-2(0+1)^2+7
y=-2(1)+7
y=-2+7
y=5

The y-intercept is (0,5)

For finding x-intercepts, we need to plug in y=0

0=-2(x+1)^2+7
Subtract 7 from both ends and isolating the term containing x
-7 = -2(x+1)^2

Let us rewrite it as -2(x+1)^2=-7 It looks better to when the variable is kept of the left side of the equation.

-2(x+1)^2=-7 dividing by -2 on both sides isolates (x+1)^2

We get

(x+1)^2=7/2

Take square root on both the sides we get

sqrt((x+1)^2) = +-sqrt(7/2)
x+1 = +-sqrt(7/2)

Subtract 1 from both sides to solve for x

x=-1+-sqrt(7/2)

The x-intercepts are (-1+sqrt(7/2)) and (-1-sqrt(7/2))