Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Mid-Chapter Quiz
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Exercise 17 Page 224

Substitute the given (x,y) values into y=ax^2+bx+c to write a system of three equations.

y=x^2+6x+3

Practice makes perfect
Before we try to write the equation of a specific parabola that passes through some given points, let's make sure we have enough points. We will need at least three. (0,3),(1,10),(2,19) To use the given points, we need to substitute their ( x, y) coordinate pairs into the standard form of a quadratic equation. y=a x^2+b x+c, a≠ 0 Doing so will create a system of equations that we can solve for the values of a, b, and c. Let's start with (0,3).
y=ax^2+bx+c
3=a( 0)^2+b( 0)+c
Simplify
3=a(0)+b(0)+c
3=0+0+c
3=c
c=3
We already found that c=3! y=ax^2+bx+3 Now we can use (1,10) to write our second equation.
y=ax^2+bx+3
10=a( 1)^2+b( 1)+3
Simplify
10=a(1)+b(1)+3
10=a+b+3
7=a+b
a+b=7
To find another equation, we will use (2,19).
y=ax^2+bx+3
19=a( 2)^2+b( 2)+3
Simplify
19=a(4)+b(2)+3
19=4a+2b+3
16=4a+2b
8=2a+b
2a+b=8
We now have a system of two equations. a+b=7 & (I) 2a+b=8 & (II) Let's solve this system using the Elimination Method. We will start by subtracting Equation (I) from Equation (II) to eliminate the b-variable.
a+b=7 2a+b=8
a+b=7 2a+b-( a+b)=8- 7
a+b=7 2a+b-a-b=8-7
a+b=7 a=1
We found that a=1. y=1x^2+bx+3 ⇔ y=x^2+bx+3 Let's now substitute 1 for a in Equation (I).
a+b=7 a=1
1+b=7 a=1
b=6 a=1
Now that we have all three values, we can complete the standard form equation of the parabola that passes through the given points. y=1x^2+6x+3 ⇔ y=x^2+6x+3 To help visualize this graph, we have plotted the given points and sketched the curve below.