**Step 2: Find equations modeling both parabolas in vertex form**

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**Vertex: (-3,29/2)**

**Other point: (-12, 5)**

**Equation:**

**5=a(-12-(-3))**^{2} +29/2

**5=a(-9)**^{2} +29/2

**-19/2=a(-9)**^{2}

**-19/2=81a**

**a=-19/162**

**y=-19/162 (x+3)**^{2}+(29/2)

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**First, find the vertex and one other point.**

### The equation of a parabola can be written either in standard form (y=ax^{2}+bx+c) or vertex form (y=a(x-h)^{2}+k, where (h,k) is the vertex).

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### In this case, we can find the vertex and one other point by mapping the photos from Step 1 onto graph paper. Then we plug the coordinates for x, y, h, and k into the general vertex form equation. We can then solve for a by combining like terms in the parentheses, bringing the constant over to the other side using the subtraction property of equality, simplifying the exponential, and dividing both sides by the coefficient of a. We can then plug in all values to get the final equation.

**Vertex: (-11/2,5) Other Point: (-14,3)**

**Equation:**

** 3=a(-14-(-11/2))**^{2}+5

**-2=a( -17/2)**^{2}

**-2=a(289/4)**

**a=-8/289**

**y=-8/289(x+5.5)**^{2}+5

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