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Substitute the given (x,y) values into y=ax^2+bx+c to write a system of three equations.
y=4x^2+5x+1
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).
x= - 2, y= 7
Calculate power
Multiply
a(- b)=- a * b
Rearrange equation
We just wrote our first equation! Now let's do the same thing for (1,10).
x= 1, y= 10
1^a=1
Identity Property of Multiplication
Rearrange equation
To find our third and last equation, we will use (2,27).
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.
(I): Subtract (II)
(I): Distribute - 1
(I): Subtract terms
Now let's subtract Equation (II) from Equation (III) to eliminate the c-variable once more.
(III): Subtract (II)
(III): Distribute - 1
(III): Subtract terms
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.
(III): Subtract (I)
(III): Distribute - 1
(III): Add and subtract terms
(III): .LHS /4.=.RHS /4.
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.
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).
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.