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saul85 [17]
3 years ago
9

Please Help me in begging

Mathematics
1 answer:
Bond [772]3 years ago
8 0
    -9(5-4v+7v)
=-9(5+3v)
=-45-27v
-45-27v is the final answer.
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The graph of which function has a minimum located at (4, –3)?
valentinak56 [21]

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) <u>but is not a minimum</u>

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

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Step-by-step explanation:

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