Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
Chapter Review
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Exercise 15 Page 86

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

y=4x^2+5x+1

Practice makes perfect

We want to write an equation for the parabola that passes through the given points. (- 2,7),(1,10),(2,27) 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 (- 2,7).

y=ax^2+bx+c
7=a( - 2)^2+b( - 2)+c
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Simplify
7=a(4)+b(- 2)+c
7=4a+b(- 2)+c
7=4a-2b+c
4a-2b+c=7

We just wrote our first equation! Now let's do the same thing for (1,10).

y=ax^2+bx+c
10=a( 1)^2+b( 1)+c
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Simplify
10=a(1)+b(1)+c
10=a+b+c
a+b+c=10

To find our third and last equation, we will use (2,27).

y=ax^2+bx+c
27=a( 2)^2+b( 2)+c
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Simplify
27=a(4)+b(2)+c
27=4a+2b+c
4a+2b+c=27

We now have a system of three equations. 4a-2b+c=7 & (I) a+b+c=10 & (II) 4a+2b+c=27 & (III) Let's solve this system using the Elimination Method. We will start by subtracting Equation (II) from Equation (I) to eliminate the c-variable.

4a-2b+c=7 a+b+c=10 4a+2b+c=27
4a-2b+c-( a+b+c)=7-( 10) a+b+c=10 4a+2b+c=27
4a-2b+c-a-b-c=7-10 a+b+c=10 4a+2b+c=27
3a-3b=- 3 a+b+c=10 4a+2b+c=27

Now let's subtract Equation (II) from Equation (III) to eliminate the c-variable once more.

3a-3b=- 3 a+b+c=10 4a+2b+c=27
3a-3b=- 3 a+b+c=10 4a+2b+c-( a+b+c)=27-( 10)
3a-3b=- 3 a+b+c=10 4a+2b+c-a-b-c=27-10
3a-3b=- 3 a+b+c=10 3a+b=17

Now, neither Equation (I) nor Equation (III) includes the c-variable. These equations form a system with only two variables, a and b. Let's solve this system by using the Elimination Method again.

3a-3b=- 3 a+b+c=10 3a+b=17
3a-3b=- 3 a+b+c=10 3a+b-( 3a-3b)=17-( -3)
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(III): Solve for b
3a-3b=- 3 a+b+c=10 3a+b-3a+3b=17+3
3a-3b=- 3 a+b+c=10 4b=20
3a-3b=- 3 a+b+c=10 b=5

We found our first value, allowing us to form a partial equation. y=ax^2+5x+c Let's substitute 5 for b in Equation (I) to find the value of a.

3a-3b=- 3 a+b+c=10 b=5
3a-3( 5)=- 3 a+b+c=10 b=5
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(I): Solve for a
3a-15=- 3 a+b+c=10 b=5
3a=12 a+b+c=10 b=5
a=4 a+b+c=10 b=5

With our second value, we can continue forming the partial equation. y=4x^2+5x+c Finally, to find the value of c, we will substitute a=4 and b=5 into Equation (II).

a=4 a+b+c=10 b=5
a=4 4+ 5+c=10 b=5
â–¼
(II): Solve for c
a=4 9+c=10 b=5
a=4 c=1 b=5

Now that we have all three values, we can complete the standard form equation of the parabola that passes through the given points. y=4x^2+5x+1 To help visualize this graph, we have plotted the given points and sketched the curve below.