How do you find the axis of symmetry and vertex point of the function: y = -x^2 - x + 9y=x2x+9?

2 Answers

Vertex: (-1/2,9 1/4)(12,914)
Line of symmetry: x=-1/2x=12

Explanation:

use -b/(2a)b2a to find the vertex:

-(-1)/(2xx(-1)) = -1/212×(1)=12

plug -1/212 in for xx

= 9 1/4=914

vertex: (-1/2,9 1/4)(12,914)

on parabolic functions the axis of symmetry is the vertical line where the vertex is, so the equation is:

x=-1/2x=12

An alternate way, using the vertex form:

Explanation:

We can change the standard form of the equation to the vertex form, which has the general form:

y=a(x-h)^2+ky=a(xh)2+k

We get there by completing the square (we do that by taking the constant of the xx term, halving it, squaring the half, then adding and subtracting it into the equation:

y=-x^2-x+9y=x2x+9

y=-(x^2+x)+9y=(x2+x)+9

y=-(x^2+x+1/4-1/4)+9y=(x2+x+1414)+9

y=-((x+1/2)^2-1/4)+9y=((x+12)214)+9

y=-(x+1/2)^2+1/4+9y=(x+12)2+14+9

y=-(x+1/2)^2+9 1/4y=(x+12)2+914

The vertex is given by (h,k)(h,k), which in this case is (-1/2, 9 1/4)(12,914).

The axis of symmetry runs through the vertex vertically and so takes the form of x= "the " x " value of the vertex"x=the x value of the vertex, which in this case is

x=-1/2x=12

This can all be seen in the graph:

graph{-x^2-x+9[-3,3,8,10]}

note that I've adjusted the graph unequally vertically and horizontally to better show the vertex