Finding the Parts of a Parabola | CK-12 Foundation - Free Printable

Let’s solve this step by step.

We are given a parabola (a U-shaped curve) on a coordinate grid. The graph opens downward, and we can see several key points marked:

- Two x-intercepts (where the graph crosses the x-axis): at (-5, 0) and (1, 0)
- The vertex (the highest point of the parabola): at (-2, 9)
- Two other symmetric points: (-4, 5) and (0, 5)

Since it’s a parabola, its equation is quadratic:
y = ax² + bx + c

But because we know the x-intercepts, we can use the factored form of a quadratic:

> y = a(x - r)(x - s)
> where r and s are the roots (x-intercepts)

From the graph, the x-intercepts are at x = -5 and x = 1.

So plug those in:

> y = a(x - (-5))(x - 1)
> → y = a(x + 5)(x - 1)

Now we need to find “a”. We can use any other point on the graph — let’s use the vertex (-2, 9), since it’s clearly labeled.

Plug x = -2, y = 9 into the equation:

> 9 = a(-2 + 5)(-2 - 1)
> 9 = a(3)(-3)
> 9 = a(-9)
> → a = 9 / (-9) = -1

So now we have:

> y = -1(x + 5)(x - 1)

Let’s expand this to standard form (optional, but good for checking):

First multiply (x + 5)(x - 1):

= x² - x + 5x - 5
= x² + 4x - 5

Now multiply by -1:

> y = -x² - 4x + 5

Let’s verify with another point — say (0, 5):

Plug x = 0:

y = -(0)² - 4(0) + 5 = 5 ✔

Check (-4, 5):

y = -(-4)² - 4(-4) + 5 = -16 + 16 + 5 = 5 ✔

Check vertex (-2, 9):

y = -(-2)² - 4(-2) + 5 = -4 + 8 + 5 = 9 ✔

Perfect! All points match.

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Final Answer:
y = -x² - 4x + 5