How do you find the vertex of #F(x)=x^2+2x-8#?

3 Answers
Apr 10, 2018

-1

Explanation:

For a quadratic function like this, the vertices are points of symmetry.

We can use a shortcut such that:
#x=(-b)/(2a)#

In your case, #b# is 2 and #a# is 1. So:
#-2/(2*1)=-1#

Confirm it on a graph.
graph{x^2+2x-8 [-25.8, 25.8, -18.1, 7.7]}

Apr 10, 2018

The vertex is when the gradient is 0.

Explanation:

Differentiate and find #x# where #F'(x) = 0#

#F(x) = x^2 + 2x - 8#

#F'(x) = 2x + 2#

#0 = 2x+ 2#

#x = -1#

#F(-1) = (-1)^2 + 2(-1) - 8#

#F(-1) = -9#

Therefore the vertex is: #(-1, -9)#

Apr 10, 2018

Vertex is (-1, -9)

Explanation:

To find the x-value of the vertex, we use the formula #x=(-b)/(2a)# (This is considered the axis of symmetry).

#a=1# and #b=2# with #c=-8# (These are the numbers in order)

By substituting we get #x=(-(2))/(2*1)#, which simplifies to #x=-1#.

Once you find x, substitute back into the equation to find y.
#y=(-1)^2+2(-1)-8#
#y=1-2-8#
#y=-9#

Putting the two numbers together you get the coordinates #(-1, -9)#.