Question

Need help finding equation

Answer 1

Answer:

Step-by-step explanation:

Based on the two graphs I see at the bottom of the photo, you're looking at a quadratic function. The answer you put in is a linear equation.

Quadratic functions follow the form $ax^2+bx+c$, where a, b, and c can be either positive or negative constants.

The table says that the equation, when graphed, should start at the top, slide to the bottom, and then slide back up to the top. So a must be positive.

c has to be -8 since it's the y-intercept.

So far, our equation is $ax^2+bx-8$. We'll need to plug in some points in order to work out what $a$ and $b$ are.

Let's use $(-2,5)$ as the first set of points, and $(4,14)$ as the second set.

$5=a(-2)^2+b(-2)-8\\14=a(4)^2+b(4)-8$

Now we have 2 equations with 2 unknowns. Solve for one variable:

$5=4a-2b-8\\13+2b=4a\\\frac{13+2b}{4}=a$

Now substitute that into the second equation:

$14=(\frac{13+2b}{4})(4^2)+4b-8\\14+8=(13+2b)(4)+4b\\22-52=8b+4b\\-30=12b\\-\frac{30}{12}=-\frac{10}{4}=-\frac{5}{2}=b$

Now substitute the result into the first equation.

$\frac{13+2(-\frac{5}{2})}{4}=a\\\\\frac{13-5}{4}=a\\\\\frac{8}{4}=a\\\\ 2=a$

Now substitute our values back into the original equation:

$y(x)=2x^2-\frac{5}{2}x-8$

Please let me know if this is correct.