Question

write the equation of the parabola that passes through the points shown in the table.

x	y
-1	0
4	2
5	0

write your answer in the form $y = a(x - p)(x - q)$, where $a$, $p$, and $q$ are integers, decimals, or simplified fractions.

Explanation:

Step1: Identify roots from table

The roots (where \( y = 0 \)) are at \( x=-1 \) and \( x = 5 \). So, \( p=-1 \), \( q = 5 \), and the equation is \( y=a(x + 1)(x - 5) \).

Step2: Find \( a \) using a point

Use the point \( (4,2) \). Substitute \( x = 4 \), \( y = 2 \) into \( y=a(x + 1)(x - 5) \):
\( 2=a(4 + 1)(4 - 5) \)
\( 2=a(5)(-1) \)
\( 2=-5a \)
\( a=-\frac{2}{5} \)? Wait, no—wait, \( 4 - 5=-1 \), \( 4 + 1 = 5 \), so \( 2=a(5)(-1)\Rightarrow a=-\frac{2}{5} \)? Wait, no, maybe miscalculation. Wait, \( x = 4 \), \( y = 2 \):
\( 2=a(4 - (-1))(4 - 5) \)? No, original form is \( y=a(x - p)(x - q) \), where \( p \) and \( q \) are roots. So roots are \( x=-1 \) (so \( p=-1 \), so \( x - p=x + 1 \)) and \( x = 5 \) (so \( q = 5 \), \( x - q=x - 5 \)). So equation is \( y=a(x + 1)(x - 5) \). Now plug in \( (4,2) \):
\( 2=a(4 + 1)(4 - 5)\Rightarrow 2=a(5)(-1)\Rightarrow 2=-5a\Rightarrow a=-\frac{2}{5} \)? Wait, that can't be. Wait, maybe I mixed up \( p \) and \( q \)? Wait, no, the roots are at \( x=-1 \) and \( x = 5 \), so the factors are \( (x + 1) \) and \( (x - 5) \). Wait, but let's check again. Wait, when \( x = 4 \), \( y = 2 \). Let's compute \( (4 + 1)(4 - 5)=5\times(-1)=-5 \). So \( 2=a\times(-5)\Rightarrow a=-\frac{2}{5} \)? But the problem says \( a \), \( p \), \( q \) can be integers, decimals, or simplified fractions. Wait, maybe I made a mistake. Wait, let's check the table again: \( x=-1 \), \( y=0 \); \( x = 4 \), \( y=2 \); \( x = 5 \), \( y=0 \). So the parabola has roots at \( x=-1 \) and \( x = 5 \), so the axis of symmetry is at \( x=\frac{-1 + 5}{2}=2 \). Wait, but when \( x = 4 \), which is 2 units to the right of the axis (since axis is at 2, 4 - 2 = 2), and \( x=-1 \) is 3 units left of axis (2 - (-1)=3). Wait, maybe using vertex form? No, the problem asks for \( y=a(x - p)(x - q) \). So let's proceed. So we have \( y=a(x + 1)(x - 5) \). Plug in \( (4,2) \):
\( 2=a(4 + 1)(4 - 5)\Rightarrow 2=a(5)(-1)\Rightarrow a=-\frac{2}{5} \). Wait, but that seems odd. Wait, maybe I messed up the roots. Wait, when \( x=-1 \), \( y=0 \); \( x = 5 \), \( y=0 \). So the quadratic is \( y=a(x + 1)(x - 5) \). Let's expand it: \( y=a(x^2 - 5x + x - 5)=a(x^2 - 4x - 5) \). Now plug in \( x = 4 \): \( y=a(16 - 16 - 5)=a(-5) \). So \( y=-5a \). But \( y = 2 \), so \( -5a=2\Rightarrow a=-\frac{2}{5} \). So the equation is \( y=-\frac{2}{5}(x + 1)(x - 5) \). Wait, but let's check with \( x = 4 \): \( -\frac{2}{5}(5)(-1)=-\frac{2}{5}\times(-5)=2 \), which matches. With \( x=-1 \): \( -\frac{2}{5}(0)(-6)=0 \), matches. With \( x = 5 \): \( -\frac{2}{5}(6)(0)=0 \), matches. So the equation is \( y=-\frac{2}{5}(x + 1)(x - 5) \). Wait, but the problem says "integers, decimals, or simplified fractions". So \( a=-\frac{2}{5} \), \( p=-1 \), \( q = 5 \). So the equation is \( y=-\frac{2}{5}(x - (-1))(x - 5) \), which is \( y=-\frac{2}{5}(x + 1)(x - 5) \).

Wait, but maybe I made a mistake in the sign. Let's rederive. The factored form of a quadratic with roots \( r_1 \) and \( r_2 \) is \( y=a(x - r_1)(x - r_2) \). So here, \( r_1=-1 \), \( r_2=5 \), so \( y=a(x - (-1))(x - 5)=a(x + 1)(x - 5) \). Then plug in \( (4,2) \): \( 2=a(4 + 1)(4 - 5)\Rightarrow 2=a(5)(-1)\Rightarrow a=-\frac{2}{5} \). So that's correct.

Answer:

\( y = -\frac{2}{5}(x + 1)(x - 5) \)