Question

A golfer aims for a hole that is 120 feet away and, when he hits the golf ball off the ground, its highest point reaches 80 feet up in the air. Noemi’s team is working together to write the quadratic equation to model the path of the golf ball in graphing form. She calculated the value of a = - 1/90 . Is she correct?

Answer 1

Answer:

No she isn't. The value of $a$ is $a=-\frac{1}{45}$

Step-by-step explanation:

We know that all quadratic equation can be written in the following way :

$y=ax^{2}+bx+c$ (I)

Where $a,b$ and $c$ are real numbers.

I will attach a drawing with the quadratic graph to understand the situation.

We know by looking at the drawing and analyzing it that the parabola passes through the points : $(0,0) ; (60,80)$ and $(120,0)$

$(0,0)$ because we put our coordinates origin there.

$(120,0)$ because that's where the golf hole is.

And $(60,80)$ because we know that its highest point reaches 80 feet up in the air at the middle of the distance between its roots (property of a negative parabola).

Finally, we work with the three points and the equation (I) in order to find the values of $a,b$ and $c$ ⇒

The parabola passes through $(0,0)$ ⇒

$y=ax^{2}+bx+c$ ⇒

$0=a(0)^{2}+b(0)+c$ ⇒ $c=0$

The parabola passes through $(60,80)$ ⇒

$80=a(60)^{2}+b(60)$ ⇒

$80=3600a+60b$ (II)

The parabola passes through $(120,0)$ ⇒

$0=a(120)^{2}+b(120)$ ⇒

$0=14400a+120b$

$120b=-14400a$

$b=-120a$ (III)

Now if we use (III) in (II) ⇒

$80=3600a+60(-120a)$

$80=3600a-7200a$

$3600a=-80$

$a=-\frac{1}{45}$    ⇒ $b=\frac{8}{3}$

Finally the equation of the parabola is

$y=-\frac{x^{2}}{45}+\frac{8}{3}x$

Where the value of $a$ is $a=-\frac{1}{45}$