Question

The graph of which function has a minimum located at (4, –3)?

Answer 1

Answer:

its c

Step-by-step explanation:

Answer 2

Answer:

None of the options is the answer to the question

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

case A) we have

$f(x)=x^{2}+4x-11$

In this case the x-coordinate of the vertex will be negative

therefore

case A is not the solution

case B) we have

$f(x)=-2x^{2}+16x-35$

This case is a vertical parabola open downward (the vertex is a maximum)

The vertex is the point $(4,-3)$ but is not a minimum

see the attached figure

therefore

case B is not the solution

case C) we have

$f(x)=x^{2}-4x+5$

Convert into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-5=x^{2}-4x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-5+4=(x^{2}-4x+4)$

$f(x)-1=(x^{2}-4x+4)$

Rewrite as perfect squares

$f(x)-1=(x-2)^{2}$

$f(x)=(x-2)^{2}+1$ --------> vertex form

The vertex is the point $(2,1)$

therefore

case C is not the solution

case D) we have

$f(x)=2x^{2}-16x+35$  

Convert into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-35=2x^{2}-16x$

Factor the leading coefficient

$f(x)-35=2(x^{2}-8x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-35+32=2(x^{2}-8x+16)$

$f(x)-3=2(x^{2}-8x+16)$

Rewrite as perfect squares

$f(x)-3=2(x-4)^{2}$

$f(x)=2(x-4)^{2}+3$ --------> vertex form

The vertex is the point $(4,3)$

therefore

case D is not the solution

The answer to the question will be the function

$f(x)=2x^{2}-16x+29$