Answer:
Option B [tex]x=-2[/tex]
Step-by-step explanation:
we know that
The critical point of a function are the points on the graph of a function where the derivative is zero or the derivative does not exist.
we have
[tex]f(x)=2x^{2}+8x-7[/tex]
step 1
Take the derivative of the function
[tex]f'(x)=4x+8[/tex]
step 2
Set that derivative equal to 0 and solve for x. Each x value you find is known as a critical number
[tex]f'(x)=0[/tex]
[tex]4x+8=0[/tex]
[tex]4x=-8[/tex]
[tex]x=-2[/tex]
Alternative Method
The critical point of the quadratic equation is the vertex, because the function changes from decreasing to increasing at that point (In this problem the vertex is a minimum)
we have
[tex]f(x)=2x^{2}+8x-7[/tex]
Convert into vertex form
[tex]f(x)+7=2x^{2}+8x[/tex]
[tex]f(x)+7=2(x^{2}+4x)[/tex]
[tex]f(x)+7+8=2(x^{2}+4x+4)[/tex]
[tex]f(x)+15=2(x^{2}+4x+4)[/tex]
[tex]f(x)+15=2(x+2)^{2}[/tex]
[tex]f(x)=2(x+2)^{2}-15[/tex]
the vertex is the point (-2,-15)
therefore
The x-coordinate of the critical point is x=-2