Question

Create a quadratic function that has a solution at (-3, 13). Demonstrate that your function has thissolution.

Answer 1

ANSWER:

$\begin{gathered} y=\sqrt[]{178-x^2} \\ or \\ y=\frac{13}{9}x^2 \end{gathered}$

STEP-BY-STEP EXPLANATION:

In the case that (-3, 13) is a point (x, y), the answer is the following:

The way to solve this problem is through trial and error.

In this case, we can propose the following quadratic equation:

$y=\sqrt{178-x^2}$

We check the equation by replacing in the function:

$\begin{gathered} y=\sqrt[]{178-(-3)^2} \\ y=\sqrt[]{178-9} \\ y=\sqrt[]{169} \\ y=\pm13 \end{gathered}$

Therefore, the point (-3, 13) is part of the solution of the proposed function

Or we can opt for the 3-point method:

For (0, 0) we have

0 = C

For (3, 13) we have

13 = 9a + 3b

For (-3, 13) we have

13 = 9a - 3b

adding them:

$\begin{gathered} 13+13=9a+9a+3x-3b \\ 26=18a \\ a=\frac{13}{9} \\ \text{for b} \\ 13=9\cdot\frac{13}{9}+3b \\ 3b=0 \\ b=0 \\ \text{the equation would it:} \\ y=\frac{13}{9}x^2 \\ \text{replacing the point (-3,13)} \\ 13=\frac{13}{9}\cdot(-3)^2 \\ 13=13 \end{gathered}$