How do you find the vertex and axis of symmetry, and then graph the parabola given by: #y=x^(2)-2x-15#?

Given:

#y=x^2-2x-15# is a quadratic equation in standard form:

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

where:

#a=1#, #b=-2#, and #c=-15#

The formula for the axis of symmetry and the #x# value of the vertex is:

#x=(-b)/(2a)#

Plug in the known values.

#x=(-(-2))/(2*1)#

Simplify.

#x=2/2=1#

To find the #y# value of the vertex, substitute #1# for #x# in the equation.

#y=1xx1^2-2(1)-15#

Simplify.

#y=1-2-15#

#y=-16#

The vertex is #(1,-16)#. Since #a>0#, the vertex is the minimum point and the parabola opens upward.

We also need the x-intercepts, which are the values of #x# when #y=0#. Substitute #0# for #y# and solve for #x# by factoring.

#0=x^2-2x-15#

Find two numbers that when added equal #-2#, and when multiplied equal #-15#. The numbers #-5# and #3# meet the requirements.

#0=(x-5)(x+3)#

#0=x-5#

#5=x#

#0=x+3#

#-3=x#

#x=5,-3#

The x=intercepts are #(5,0)# and #(-3,0)#.

Summary

The axis of symmetry is #1#.

The vertex is #(1,-16)#

The x-intercepts are #(5,0)# and #(-3,0)#.

Now plot the vertex and the x-intercepts. Sketch a parabola through the points. Do not connect the dots.

graph{x^2-2x-15 [-14.82, 17.2, -16.08, -0.06]}