# Thread: Quadratic Function,HOW TO FIND SPECIFIC VALUE'S ON GRAPH USING FORMULA

1. ## Quadratic Function,HOW TO FIND SPECIFIC VALUE'S ON GRAPH USING FORMULA

I need help asap.

Both questions refer to the quadratic function y=-0.7x2+3x+0.2+1.78
I have also attached the graph of this function, using DESMOS, it is attached as an .HTML file.

(BTW The equation refers to 'the perfect shot equation'. In basketball the perfect shot equation follows a quadratic function, wherein the ball thrown goes in every time. The data on the graph is in meters where the y-axis represents height and x-axis distance.

THE QUESTIONS I CANT FIGURE OUT:
"a) If the height of the basketball ring is 3m, show how you can use your equation to determine how far in from the baseline it is?"

!?!?

"b) Now you know the quadratic equation to shoot the perfect shot, but where do you need to stand? Investigate where your 'sweet spot' (the spot where your shot will follow the equation and go in every time) on the court is. Explain how you derived your answer. (Hint: use your graph to determine the distance from the basketball ring you need to be and check it using your equation)"

!?!?

Thanks in advance to any and everybody!

2. ## Re: Quadratic Function,HOW TO FIND SPECIFIC VALUE'S ON GRAPH USING FORMULA

For (a), there must be some other data provided that you haven't shared. For example, is it assumed that with the "perfect shot" the ball passes through the hoop and hits the floor precisely on the baseline?
For (b), set y = 3 meters and solve for x. There are two solutions for this: one when the ball is on its way up and the other when it's on its way down.

3. ## Re: Quadratic Function,HOW TO FIND SPECIFIC VALUE'S ON GRAPH USING FORMULA

Yes for a) the x intercept is the baseline,

can you explain b) a little in more depth

Thanks for the quick reply mate!

4. ## Re: Quadratic Function,HOW TO FIND SPECIFIC VALUE'S ON GRAPH USING FORMULA

So for (a) set y = 0 and calculate what x is. You can do this either by using the quadratic formula, or you can "eye ball" it from the graph you have. You should have the quadratic equation memorized:the solution for ax^2 + bx + c = 0 is:

$x = \frac {-b \pm \sqrt{b^2-4ac}}{2a}$

For (b), you can also solve using the quadratic equation, this time with y = 3m. In other words, solve: 3 = -0.7x^2+3x + 0.2 + 1.78. Rearranging yields 0 = -0.7x^2 +3x -1.02, which again can be solved using the quadratic equation. Or, determine it from your graph.