Select the x-coordinate of the vertex......

Question

Select the x-coordinate of the vertex of the parabola defined by the function f(x)=-9x^2 +5x+7?

a) - 9/5

b) 7

c) 5/18

d) - 5/9

Answer 1

When you have a quadratic polynomial (ax² + bx + c) that represents a parabola, you can find the x-coordinate of the vertex using the formula x=-b/(2a).

 

In your case, a=-9, and b=5. Can you figure out the x-value now?

Answer 2

f(x)=-9x^2 +5x+7

 

Factor out -9 from the first two terms...

 

-9(x^2 -(5/9)x ) + 7

 

Complete the square by multiplying -5/9 by 1/2 and squaring it. (-5/18)^2 = 25/324

 

y = -9(x^2 - (5/9)x + 25/324) + 7 + 9(25/324) ...

Since you added -9(25/324) inside,  add +9(25/324)  outside to have a net change of 0 on the right side of the equation.

 

9(25/324) = (25/36)

 

7 = 252/36

 

y = -9(x^2 - (5/9)x +25/324) + 252/36 + 25/36

 

y = -9(x - 5/18)^2 + 277/36

 

vertex = (5/18, 277/36)  

 

277/36 is approximately 7.694

 

 

 

So Byron is right, as we know from the shortcut.  Just do it HIS way!  :)

I just wanted the student to see why it works, and why he shouldn't do it by completing the square, unless the teacher specifies it must be done that way.  Still use the shortcut to check your final answer.

 

 

Notice that in completing the square we took b which was 5, made it negative,  and divided by [   2(same as multiplying by 1/2) * -9 (which is a)   ].

 

-5

===

2(-9)

 

So x = -b/(2a), just like Byron stated.

Answer 3

y=-9x²+5x+7

Let u=3x

y=-u²+(5/3)u+7=-(u²-(5/3)u-7)

y=-[(u-5/6)²-25/36-7]=-[(u-5/6)²-277/36]

y-277/36=-(3x-5/6)²=-(3(x-5/18)²=9(x-5/18)²

(y-277/36)/9=-(x-5/18)²

The vertex is at (5/18,277/36), the 1/9 represents 4 times the focal length.

Thank you Jason  for the sign correction.

 

Answer 4

I retract my opinion.  Byron, you're right.  Please accept my apology.