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The question also states that the parabola passes through points $(-3,17)$, $(3,17)$, and $(2,7)$.

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    What do you know? have you tried anything? – 2017-01-07
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    You could let the equation of the parabola be $f(x)=ax^2+bx+c$, then sub in the individual points to get a series of 3 equations. Then solve the 3 simultaneous equations to obtain the values of $a,b,c$. – 2017-01-07
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    Well, I tried putting them in in standard form for some reason, but I think I did something wrong, because after some of the steps I have done, I got a as a long decimal. – 2017-01-07

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Parabolas that are functions have a vertical line of symmetry. Because your parabola fits f(-3) = f(3), the parabola has the line of symmetry x=0.

Therefore, it is of the form $y = ax^2 + b$. We can plug in x=2 and x=3 to find the respective points.

Plugging in 2 gives us 7 = 4a + b. (1)

Plugging in 3 gives us 17 = 9a + b. (2)

Subtracting (1) from (2) gives us 10 = 5a. Therefore, a=2.

Plugging in a=2 into (1) gives us 7 = 8 + b. Therefore, b = -1.

Plugging this into our equation gives us $y = 2(x^2)-1$.

If you don't want to go through this method, try plugging in all three points into the equation $y = ax^2 + bx + c$. It will yield the same $a = 2$, $b = 0$, $c = -1$.

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Because the two first points are symmetric, the component X is 0. Now, the equation for this parabola is $2 x^2-1$ and the y component is - 1

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Let $y=f(x)=ax^2+bx+c$

If $(-3,17), (3,17)$, and $(2,7) \in y=f(x)$. Then $$f(-3)=17$$ $$f(3)=17$$ $$f(2)=7$$ Then $$a=..., b=..., c=...$$ Answer: $$x_0=-\frac b{2a}$$

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Hint:

Since the parabola has identical $y$ values at $x=-3$ and $x=3$, the vertex must be on the line $x=0$.

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    With this, how do I find the vertex, though? I know that in standard form, one would use x = -b/2a, then sub that result to get the y-coordinate, but I'm unsure about this one. – 2017-01-07
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    Well, you know that the parabola has equation $y=ax^2+bx+c$. Since the vertex is at $x=0$, you know that $b=0$. Now, all you need to do is plug in 2 of your coordinates into $y=ax^2+c$ and solve the simultaneous equations to evaluate the values of $a$ and $c$. – 2017-01-07
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    But, how come b = 0 when the vertex's x-coordinate is 0? I'm not sure why. – 2017-01-07
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    An alternative equation of a parabola is $y=a(x-h)^2+k$ where $(h,k)$ are the coordinates of the vertex. If $h=0$, the form of the equation is $ax^2+k$, which means that the coefficient of $x$ in your parabola is $0$. Hence, $b=0$ in $y=ax^2+bx+c$. – 2017-01-07
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As an alternative approach, given the value of a quadratic function at three distinct points you can find the $x^2$ term almost directly.

First, we compute the slopes of two of the secant lines connecting pairs of given points. Usually I would choose consecutive pairs of points from left to right but in this example I will use the pair $(−3,17)$ and $(3,17)$ for one of the lines and the pair $(2,7)$ and $(3,17)$ for the other because the slopes of these lines are so easy to compute. The line through $(−3,17)$ and $(3,17)$ has slope $0$ and the line through $(2,7)$ and $(3,17)$ has slope $10.$

Now a fact about parabolas, which may seem like an application of calculus but actually can be derived from facts known in ancient times, is that if we take the $x$ coordinate at the midpoint of a secant line between two points on the parabola, the tangent to the parabola at the same $x$ coordinate is parallel to the secant line. So we know that at $x=0,$ the parabola we're looking for has slope $0.$ (This incidentally tells us that the vertex is at $x=0$.) We also know that at $x=2.5,$ the parabola has slope $10.$

A second fact about parabolas (also derivable from facts known in ancient times, not just through calculus) is that the difference in the slope of the tangent line at two points of the parabola is the difference in their $x$ coordinates multiplied by $2a.$ In this example, (where the slope increases by $10$ when we go from $x=0$ to $x=2.5$), this implies that $$ 10 = 2.5 \times 2a. $$ We can easily solve for $a,$ finding that $a = 2.$

This tells us that the parabola has the form $y=2x^2 + c.$ We need merely plug in $x$ and $y$ from one of the given points to find the value of $c.$