# What is the ordered vertex pair, axis of symmetry, x-intercept, and y intercept for this problem f(x) = x^2 + 6x-16?

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Let y = f(x)

=`x^2+6x-16`

=`x^2+2*x*3*+3^2-25`

=`(x+3)^2-5^2`

`rArr (y+25)=(x+3)^2`

This is the equation of a parabola, X^2=4aY (where, X=(x+3) and Y=(y+25))

The vertex being at X=0, Y=0,

Thus, x+3=0, `rArr` x=-3,

and

y+25=0, `rArr` y=-25

Therefore the vertex is at (-3, -25).

Axis of symmetry may be obtained by putting X=0,

`rArr` x+3=0

`rArr` x=-3

This is the equation of the axis of symmetry of the parabola.

The x-intercept is obtained by putting y=0,

Thus `0+25=(x+3)^2`

`rArr` `(x+3)^2-5^2=0`

`rArr` (x+3-5)(x+3+5)=0

`rArr` (x-2)(x+8)=0

x=2, -8

So, there will be two x intercepts, 2 units and 8 units.

The y-intercept is obtained by putting x=0,

`y+25=(0+3)^2=9`

`rArr` y=-16

Therefore, y-intercept will be 16 units.