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

From the table, when \( y = 0 \), \( x=-1 \) and \( x = 5 \). So the roots are \( p=-1 \), \( q = 5 \). The factored form is \( y=a(x + 1)(x - 5) \).

Step2: Substitute a point to find \( a \)

Use the point \( (4,2) \). Substitute \( x = 4 \), \( y = 2 \) into \( y=a(x + 1)(x - 5) \):
\( 2=a(4 + 1)(4 - 5) \)
\( 2=a(5)(-1) \)
\( 2=-5a \)
Solve for \( a \): \( a=\frac{2}{-5}=-\frac{2}{5} \)? Wait, no, wait: \( 4 - 5=-1 \), \( 4 + 1 = 5 \), so \( 2=a(5)(-1)\Rightarrow 2=-5a\Rightarrow a=-\frac{2}{5} \)? Wait, but let's check again. Wait, maybe I made a mistake. Wait, the point is \( (4,2) \). Let's re - calculate:

\( y=a(x + 1)(x - 5) \)

Substitute \( x = 4 \), \( y = 2 \):

\( 2=a(4 + 1)(4 - 5) \)

\( 2=a(5)(-1) \)

\( 2=-5a \)

\( a=-\frac{2}{5} \)? Wait, but that seems odd. Wait, maybe I mixed up the roots. Wait, the roots are when \( y = 0 \), so \( x=-1 \) and \( x = 5 \), so the factors are \( (x+1) \) and \( (x - 5) \). Let's check with \( x = 4 \):

\( (4 + 1)=5 \), \( (4 - 5)=-1 \), product is \( -5 \). Then \( a\times(-5)=2\Rightarrow a=-\frac{2}{5} \). Wait, but let's verify with the other root. When \( x=-1 \), \( y=a(0)(-6)=0 \), correct. When \( x = 5 \), \( y=a(6)(0)=0 \), correct. When \( x = 4 \), \( y=-\frac{2}{5}(5)(-1)=-\frac{2}{5}\times(-5)=2 \), correct. So the equation is \( y=-\frac{2}{5}(x + 1)(x - 5) \). Wait, but the problem says \( y=a(x - p)(x - q) \). So if \( p=-1 \), then \( (x - p)=(x+1) \), and \( q = 5 \), so \( (x - q)=(x - 5) \). So the equation is \( y=-\frac{2}{5}(x+1)(x - 5) \). Wait, but let's re - express:

Wait, maybe I made a mistake in the sign of the roots. Let's recall that the factored form is \( y=a(x - r_1)(x - r_2) \), where \( r_1 \) and \( r_2 \) are the roots. So if the roots are \( x=-1 \) and \( x = 5 \), then \( r_1=-1 \), \( r_2 = 5 \), so \( y=a(x-(-1))(x - 5)=a(x + 1)(x - 5) \), which is what we had. Then substituting \( (4,2) \) gives us \( a=-\frac{2}{5} \). Wait, but let's check the calculation again:

\( 2=a(4 + 1)(4 - 5) \)

\( 2=a\times5\times(-1) \)

\( 2=-5a \)

\( a=-\frac{2}{5} \)

Yes, that's correct. So the equation is \( y=-\frac{2}{5}(x + 1)(x - 5) \)

Wait, but maybe I messed up the root values. Wait, the table has \( x=-1,y = 0 \); \( x = 5,y = 0 \); \( x = 4,y = 2 \). So the parabola crosses the x - axis at \( x=-1 \) and \( x = 5 \), so the factored form is \( y=a(x + 1)(x - 5) \), and then we use the point \( (4,2) \) to find \( a \).

Answer:

\( y=-\frac{2}{5}(x + 1)(x - 5) \) (or in the form \( y=a(x - p)(x - q) \), it can be written as \( y=-\frac{2}{5}(x-(-1))(x - 5) \), so \( a =-\frac{2}{5} \), \( p=-1 \), \( q = 5 \))