select the x coordinate of the vertex of the parabola defined by the functions f(X)=-7X^2+3X+9. a) -7/3 ,B) 9, c) 3/14 d) -3/7

Question

I do not know where to start when it comes to solving this equation.

Answer 1

f(x) = -7x^2 + 3x + 9 <== Standard Form

 

f(x) = -7(x-h)^2 + k <== Vertex Form, b/c Vertex = (h,k)

 

Let's multiply the Vertex Form out and compare to our Standard Form:

 

f(x) = -7(x^2-2hx+h^2) + k

 

f(x) = -7x^2 + 14hx - 7h^2 + k

f(x) = -7x^2 +    3x + 9

 

So that means 14h = 3, or h = 3/14, which is answer c.

 

Not needed here, but see that 9 = -7h^2 + k = -7(3/14)^2 + k,

 

so k = 9 + 7(3/14)(3/14) =

9*28/28 + (3/2)(3/14) =

9(28+1)/28 =

9*29/28 =

261/28

Answer 2

Hi Karri;

f(X)=-7X²+3X+9

When calculating the vertex, you are also calculating the point at which the change of slope is zero.  Henceforth, let's take the first derivative of the equation and set this equal to zero.

0=-14x+3

Let's subtract 3 from both sides...

-3+0=-14x+3-3

-3=-14x

Let's divide both sides by -14...

-3/-14=(-14x)/-14

3/14=x

This corresponds to the equation...

vertex=-b/2a of f(X)=-7X²+3X+9

ANSWER IS C

Answer 3

f( X ) = -7 X^2 + 3X + 9

 

 

  for : f( x ) = a X^2 + bX + c

 

    X _{vertex = - b/2a}

 

      X _{vertex = -3/ -14 = 7/14}

 

 

         answer C is correct