Question

1. directions

drag and drop the correct answer choice to each answer blank. or, touch the answer choice followed by the
find the quadratic equation for the table below:

x	y
-3	58
-1	22
2	13
3	22
5	58

if $y = ax^2 + bx + c$, what are the values of a, b and c?
answer choices: -6, 3, 13, 6, -13, -3

Explanation:

Step1: Observe symmetry

Notice that when \( x = -3 \) and \( x = 5 \), \( y = 58 \); when \( x = -1 \) and \( x = 3 \), \( y = 22 \). The axis of symmetry is \( x=\frac{-3 + 5}{2}=\frac{-1+3}{2}=1 \)? Wait, no, wait: the midpoint of -3 and 5 is \( \frac{-3 + 5}{2}=1 \)? Wait, no, \( x=\frac{-3 + 5}{2}=1 \)? Wait, but when \( x = 2 \), \( y = 13 \), \( x = 3 \), \( y = 22 \), \( x=-1 \), \( y = 22 \), \( x=-3 \), \( y = 58 \), \( x = 5 \), \( y = 58 \). Wait, actually, the axis of symmetry is \( x=\frac{-3 + 5}{2}=1 \)? Wait, no, let's check the x - values: -3, -1, 2, 3, 5. Wait, the symmetric points: (-3,58) and (5,58) have midpoint \( x=\frac{-3 + 5}{2}=1 \); (-1,22) and (3,22) have midpoint \( x=\frac{-1 + 3}{2}=1 \). So the vertex is at \( x = 1 \). For a quadratic \( y=ax^{2}+bx + c \), the axis of symmetry is \( x=-\frac{b}{2a}=1 \), so \( b=-2a \).

Step2: Substitute points

Let's use the point \( (2,13) \), \( (-1,22) \), and \( (3,22) \). Let's take \( x = 2 \), \( y = 13 \): \( 13=a(2)^{2}+b(2)+c=4a + 2b + c \). Take \( x=-1 \), \( y = 22 \): \( 22=a(-1)^{2}+b(-1)+c=a - b + c \). Take \( x = 3 \), \( y = 22 \): \( 22=a(3)^{2}+b(3)+c=9a+3b + c \).

Since \( b=-2a \), substitute \( b=-2a \) into the equations.

First equation: \( 13 = 4a+2(-2a)+c=4a-4a + c=c \). Wait, that's nice! So \( c = 13 \).

Now, substitute \( c = 13 \) and \( b=-2a \) into the second equation: \( 22=a-(-2a)+13=a + 2a+13=3a + 13 \). Then \( 3a=22 - 13 = 9 \), so \( a = 3 \). Then \( b=-2a=-6 \).

Let's check with \( x=-3 \): \( y=3(-3)^{2}-6(-3)+13=3(9)+18 + 13=27+18+13=58 \), which matches. \( x = 5 \): \( 3(25)-6(5)+13=75 - 30+13=58 \), matches. \( x=-1 \): \( 3(1)-6(-1)+13=3 + 6+13=22 \), matches. \( x = 3 \): \( 3(9)-6(3)+13=27-18 + 13=22 \), matches. \( x = 2 \): \( 3(4)-6(2)+13=12-12 + 13=13 \), matches. So \( a = 3 \), \( b=-6 \), \( c = 13 \).

Wait, let's do it step by step with substitution.

We have the quadratic equation \( y=ax^{2}+bx + c \).

Take the point \( (2,13) \): \( 13 = 4a+2b + c \) --- (1)

Take the point \( (-1,22) \): \( 22=a - b + c \) --- (2)

Take the point \( (3,22) \): \( 22=9a+3b + c \) --- (3)

Subtract equation (2) from equation (3): \( (9a + 3b + c)-(a - b + c)=22 - 22 \)

\( 8a+4b=0 \) => \( 2a + b=0 \) => \( b=-2a \) --- (4)

Subtract equation (2) from equation (1): \( (4a + 2b + c)-(a - b + c)=13 - 22 \)

\( 3a+3b=-9 \) => \( a + b=-3 \) --- (5)

Substitute \( b=-2a \) into equation (5): \( a+(-2a)=-3 \) => \( -a=-3 \) => \( a = 3 \)

Then from equation (4), \( b=-2(3)=-6 \)

Substitute \( a = 3 \), \( b=-6 \) into equation (2): \( 22=3-(-6)+c \) => \( 22=9 + c \) => \( c=13 \)

So \( a = 3 \), \( b=-6 \), \( c = 13 \)

Step3: Verify with other points

Check \( x=-3 \): \( y=3(-3)^{2}-6(-3)+13=3*9 + 18+13=27 + 18+13=58 \), which matches.

Check \( x = 5 \): \( y=3(5)^{2}-6(5)+13=3*25-30 + 13=75-30 + 13=58 \), which matches.

Answer:

\( a = 3 \), \( b=-6 \), \( c = 13 \)