Question

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

f(x) = x2 + 4x – 11
f(x) = –2x2 + 16x – 35
f(x) = x2 – 4x + 5
f(x) = 2x2 – 16x + 35

Answer 1

The function that has a minimum located at (4, -3) is f(x) = x² - 4x + 5.

Step-by-step explanation:

The function that has a minimum located at (4, -3) is f(x) = x² - 4x + 5.

To determine this, we need to find the vertex of each function by using the formula x = -b/(2a), where a is the coefficient of the quadratic term, and b is the coefficient of the linear term.

For f(x) = x² - 4x + 5, we have a = 1 and b = -4. Plugging these values into the formula, we get x = -(-4)/(2*1) = 4. Substituting this value of x back into the function, we find f(4) = 4² - 4(4) + 5 = -3.

Answer 2

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`