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

Question

Isaiah Anthony · Accepted Answer

Explanation is given below.

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))#

Eva Adamson · Answer

undefined To find the vertex and intercepts for the quadratic function $y = -2(x+1)^2 +7$:

1. **Vertex:** Use the vertex form of a quadratic function, which is $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex.
   - In the given function, $h = -1$ and $k = 7$, so the vertex is at $(-1, 7)$.

2. **x-intercept:** Set $y = 0$ and solve for $x$.
   - $0 = -2(x+1)^2 +7$
   - $2(x+1)^2 = 7$
   - $(x+1)^2 = \frac{7}{2}$
   - $x+1 = \pm \sqrt{\frac{7}{2}}$
   - $x = -1 \pm \sqrt{\frac{7}{2}}$

3. **y-intercept:** Set $x = 0$ and solve for $y$.
   - $y = -2(0+1)^2 +7$
   - $y = -2(1)^2 +7$
   - $y = -2 + 7$
   - $y = 5$

So, the vertex is $(-1, 7)$, the x-intercepts are $-1 + \sqrt{\frac{7}{2}}$ and $-1 - \sqrt{\frac{7}{2}}$, and the y-intercept is $5$.