How to write a quadratic function given the y-intercept and a "passing through" point?

Question

My math homework says "Write an equation for a quadratic function whose graph satisfies the given condition. Use whatever form is most convenient."

 

Number 5 says : Has y-intercept = 4 and passing through (5,1)

Number 9 says : has y-intercept = 0 and passing through (6,0)

Answer 1

Hello, thank you for taking the time to post your question!

The general idea on this type of question is that you are trying to set up an equation in the format y = ax^2 + bx + c, since it’s after a quadratic functional form to represent each of these.

On the first on since it says the y-intercept is “4”, that would be the value of c, so you can simplify it down to

y = ax^2 + bx + 4

from there then you can using the passing through point of (5,1) to figure out what the values of a and b should be

-3 = 25a + 5b

Say b = 1, that becomes

-3 = 25a + 5(1)

-8 = 25a

a = -8/25

so a valid functional form here is f(x) = -8/25x^2 + x + 4. The idea here is that there will be many functional forms possible … anything that maintains that relationship between a and b will work

I hope that helps get you moving in the right direction! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

Assume y = ax^2 + bx + c.

 

When y = 4, x = 0 (y-intercept) so c = 4.

 

Substitute (5,1) and get 1 = a(5^2) + b(5) + 4.

1 = 25a +5b +4

-3 = 25a + 5b.

 

Now suppose that a = 1. You can do this because lots of quadratics are possible to satisfy the given points (5,1) & (0,4).  You just have to write SOME equation consistent with these facts.

 

Therefore -3 = 25 + 5b;  -28 = 5b;  b = -28/5.

 

Now that's the answer, with a = 1, b = -28/5 and c = 4 substituted into y = ax^2 + bx + c.  Obviously other answers are possible, because it takes three points to determine a particular parabola.