Answer:
Part A) The function written in vertex form is [tex]f(x)=15(x+2)^{2}-79[/tex]
Part B) The vertex of the function is the point [tex](-2,-79)[/tex]
Step-by-step explanation:
Part A) Write the function in vertex form
we know that
The equation of a vertical parabola in vertex form is equal to
[tex]y=a(x-h)^{2}+k[/tex]
where
(h,k) is the vertex of the parabola
In this problem we have
[tex]f(x)=15x^{2}+60x-19[/tex]
Convert to vertex form
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]f(x)+19=15x^{2}+60x[/tex]
Factor the leading coefficient
[tex]f(x)+19=15(x^{2}+4x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]f(x)+19+60=15(x^{2}+4x+4)[/tex]
[tex]f(x)+79=15(x^{2}+4x+4)[/tex]
Rewrite as perfect squares
[tex]f(x)+79=15(x+2)^{2}[/tex]
[tex]f(x)=15(x+2)^{2}-79[/tex] -----> function in vertex form
Part B) Name the vertex for the function
we have
[tex]f(x)=15(x+2)^{2}-79[/tex]
The vertex of the function is the point [tex](-2,-79)[/tex]
The parabola open upward, so the vertex is a minimum
see the attached figure to better understand the problem
