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

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Answer 1 · 2018-03-22T14:47:26

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

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]}