Select the x coordinate of the vertex of the parabola defined by the function: f(x)=-5x^2+3x+2

A. -5/3

B. 2

C. 3/10

D. -3/5

I keep coming up with 0. I know I have missed a step, please help.

Select the x coordinate of the vertex of the parabola defined by the function: f(x)=-5x^2+3x+2

A. -5/3

B. 2

C. 3/10

D. -3/5

I keep coming up with 0. I know I have missed a step, please help.

Tutors, please sign in to answer this question.

Woodland Hills, CA

Without calculus, the Horizontal Coordinate of a quadratic of f(x) =aX^2 + bx + c is equal to - b/ 2a

Which in this case of

f( X) = - 5 X^2 + 3X + 2

equals : - 3/[2(-5)] = -3/ -10 = 3/10

Middletown, CT

Hi Christine;

f(x)=-5x^{2}+3x+2

The vertex is the point at which the change of y is zero. Therefore, we can take the first derivative of the function and calculate this...

0=-10x+3

-3=-10x

3/10=x

ANSWER C

Dupont, WA

If you have a graphing calculator you can visually see that the answer is C.

There is a formula derived from the quadratic formula for the x coordinate of the vertex which is -b/(2a)

For any parabola it will give you the x coordinate of the vertex.

f(x)=-5x^2+3x+2

a=-5, b=3, c=2

-3/(2*-5) = 3/10

The answer is C. 3/10

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