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Interpreting Quadratic Functions in Factored Form

Interpreting Quadratic Functions in Factored Form 1.10 - Solution

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We want to write the equation of the quadratic function that passes through the points (-3,0),(\textcolor{darkorange}{\text{-} 3},0), (1,-72),(1,\text{-} 72), and (4,0).({\color{#FF0000}{4}},0). Note that the y-y\text{-}coordinate of two of the points is 0.0. Therefore, we know that -3\textcolor{darkorange}{\text{-} 3} and 4{\color{#FF0000}{4}} are the x-x\text{-}intercepts of the parabola.

x\bm{x} y\bm{y}
-3\textcolor{darkorange}{\text{-}3} 00
11 -72\text{-}72
4{\color{#FF0000}{4}} 00
Consider the factored form of a quadratic function. y=a(xp)(xq)\begin{gathered} y=a(x-\textcolor{darkorange}{p})(x-{\color{#FF0000}{q}}) \end{gathered} In this form, p\textcolor{darkorange}{p} and q{\color{#FF0000}{q}} are the x-x\text{-}intercepts. Therefore, we know that, for our equation, it is p\textcolor{darkorange}{p} == -3\textcolor{darkorange}{\text{-} 3} and q{\color{#FF0000}{q}} == 4.{\color{#FF0000}{4}}. y=a(x(-3))(x4)      y=a(x+3)(x4)\begin{gathered} y=a(x-(\textcolor{#ff8c00}{\text{-}3}))(x-{\color{#FF0000}{4}}) \ \ \ \Leftrightarrow \ \ \ y=a(x+3)(x-4) \end{gathered} Since the parabola passes through the point (1,-72),(1,\text{-}72), we can substitute 11 for xx and -72\text{-}72 for yy in our partial equation, and solve for a.a.
y=a(x+3)(x4)y=a(x+3)(x-4)
-72=a(1+3)(14){\color{#009600}{\text{-}72}}=a({\color{#0000FF}{1}}+3)({\color{#0000FF}{1}}-4)
Solve for aa
-72=a(4)(-3)\text{-}72=a(4)(\text{-}3)
-72=a(-12)\text{-}72=a(\text{-}12)
-72-12=a\dfrac{\text{-}72}{\text{-}12}=a
7212=a\dfrac{72}{12}=a
6=a6=a
a=6a=6
Knowing that a=6,a=6, we can write the full equation of the parabola. y=6(x+3)(x4)\begin{gathered} y=6(x+3)(x-4) \end{gathered}