## Question

###### How do you use the important points to sketch the graph of #y = x^2 – 6x – 16#?

## Answers

Use the roots ( x = -2 and x = 8) and the vertex (3, -25) to start your sketch

## Explanation:

We know that

#y=x^2-6x-16# represents an upwardly facing parabola because the coefficient in in front of the

#x^2# term is positive (#+1# ).Finding where the parabola intersects the x-axis (if it does) will be helpful.

Find when

#y=0# .

#x^2-6x-16=0# Factor.

#(x-8)(x+2)=0# This gives two roots, namely

#x=8# and#x=-2# , so the parabola intersects the x-axis at#x=8# and#x=-2# .The x-coordinate,

#x_v# , of the vertex of the parabola is the average of these two roots.

#x_v=(-2+8)/2=3# Here's another (better) way to find

#x_v# . If we have a quadratic equation in standard form

#y=ax^2+bx+c# then

#x_v=-b/(2a)#

always. In this case,#-b=-(-6)=6# and#2a=2*1=2# so#x_v=6/2=3# .To find the y-coordinate of the vertex,

#y_v# we can substitute#x_v# into our original equation.

#y_v=x_v^2-6x_v-16# Since

#x_v=3#

#y_v=3^2-6*3-16=-25# so the vertex is at (3, -25).

Now we are ready to plot an upwardly facing parabola whose vertex is at (3, -25) and crosses the x-axis at -2 and 8.

graph{x^2-6x-16 [-5, 10,-30, 20]}