Guided Learning Extension
Write a quadratic equation from three coordinates
If you are given intercepts or the vertex, you can use intercept form or vertex form to find the equation of a quadratic function. But if you have three coordinates that aren't the intercepts or the vertex, you need to make a system of equations to solve for \(a\), \(b\) and \(c\) in standard form.
Example 1: Find the equation of the quadratic function through the following points: \((-2, -12)\), \((3, 3)\), \((2, 8)\).
\(\begin{align}&\begin{cases} -12&=4a-2b+c \\ 3&=9a+3b+c \\ 8&=4a+2b+c \end{cases}\ & \ \ \ &\text{1) Substitute each coordinate in for x and y in the equation} y=ax^2+bx+c.\\
&-15=-5a-5b\ & \ \ \ &\text{2) Combine the first two equations to eliminate c.}\\
&-5=5a+b\ & \ \ \ &\text{3) Combine the second two equations to eliminate c.}\\
&-20=-4b\ & \ \ \ &\text{4) Combine the 4th and 5th equations to eliminate a.}\\
&b=5\ & \ \ \ &\text{5) Divide to solve for b.}\\
&-5=5a+5\ & \ \ \ &\text{6) Substitute 5 for b into the fifth equation (step 3).}\\
&-10=5a\ & \ \ \ &\text{7) Subtract 5 from both sides.}\\
&-2=a\ & \ \ \ &\text{8) Divide to solve for a.}\\
&8=4(-2)+2(5)+c\ & \ \ \ &\text{9) Substitute -2 and 5 in for a and b into the third equation.}\\
&8=-8+10+c\ & \ \ \ &\text{10) Simplify.}\\
&8=2+c\ & \ \ \ &\text{11) Simplify.}\\
&6=c\ & \ \ \ &\text{12) Subtract to solve for c.}\\
&y=-2x^2+5x+6\ & \ \ \ &\text{13) Write the quadratic equation with -2, 5 and 6 substituted in for a, b, and c.}\end{align}\)
Write a quadratic equation from two non-real solutions
One way to write equations from two solutions would be to put the solutions into intercept form and distribute. This can be complicated with imaginary numbers and radicals. An easier way is to use the sum and product of roots rules that we learned in Target D. As a refresher, \(-\large\frac{b}{a}\) = sum of roots and \(\large\frac{c}{a}\) = product of roots.
Example 2: Write a quadratic function, in standard form, with the given zeros: \(1+\large\frac{\sqrt{5}}{2}i\) and \(1-\large\frac{\sqrt{5}}{2}i\). Be sure all coefficients are integers.
\(\begin{align}&x^2 - (1+\frac{\sqrt{5}}{2}i+1-\frac{\sqrt{5}}{2}i)x+(1+\frac{\sqrt{5}}{2}i)(1-\frac{\sqrt{5}}{2}i)\ & \ \ \ &\text{1) Substitute the sum of the roots in to the quadratic equation}\\
&\ & \ \ \ &\text{for b and the product of the roots in for c.}\\
&x^2-2x+1-\frac{5}{4}i^2\ & \ \ \ &\text{2) Simplify.}\\
&x^2-2x+1+\frac{5}{4}\ & \ \ \ &\text{3) Evaluate}\ \ i^2.\\
&x^2-2x+\frac{9}{4}\ & \ \ \ &\text{4) Simplify.}\\
&y=4x^2-8x+9\ & \ \ \ &\text{5) To get integers, distribute 4 throughout the equation.}\end{align}\)
Guided Learning
Write a quadratic equation from three coordinates
If you are given intercepts or the vertex, you can use intercept form or vertex form to find the equation of a quadratic function. But if you have three coordinates that aren't the intercepts or the vertex, you need to make a system of equations to solve for \(a\), \(b\) and \(c\) in standard form.
Example 1: Find the equation of the quadratic function through the following points: \((-2, -12)\), \((3, 3)\), \((2, 8)\).
\(\begin{align}&\begin{cases} -12&=4a-2b+c \\ 3&=9a+3b+c \\ 8&=4a+2b+c \end{cases}\ & \ \ \ &\text{1) Substitute each coordinate in for x and y in the equation} y=ax^2+bx+c.\\
&-15=-5a-5b\ & \ \ \ &\text{2) Combine the first two equations to eliminate c.}\\
&-5=5a+b\ & \ \ \ &\text{3) Combine the second two equations to eliminate c.}\\
&-20=-4b\ & \ \ \ &\text{4) Combine the 4th and 5th equations to eliminate a.}\\
&b=5\ & \ \ \ &\text{5) Divide to solve for b.}\\
&-5=5a+5\ & \ \ \ &\text{6) Substitute 5 for b into the fifth equation (step 3).}\\
&-10=5a\ & \ \ \ &\text{7) Subtract 5 from both sides.}\\
&-2=a\ & \ \ \ &\text{8) Divide to solve for a.}\\
&8=4(-2)+2(5)+c\ & \ \ \ &\text{9) Substitute -2 and 5 in for a and b into the third equation.}\\
&8=-8+10+c\ & \ \ \ &\text{10) Simplify.}\\
&8=2+c\ & \ \ \ &\text{11) Simplify.}\\
&6=c\ & \ \ \ &\text{12) Subtract to solve for c.}\\
&y=-2x^2+5x+6\ & \ \ \ &\text{13) Write the quadratic equation with -2, 5 and 6 substituted in for a, b, and c.}\end{align}\)
Write a quadratic equation from two non-real solutions
One way to write equations from two solutions would be to put the solutions into intercept form and distribute. This can be complicated with imaginary numbers and radicals. An easier way is to use the sum and product of roots rules that we learned in Target D. As a refresher, \(-\large\frac{b}{a}\) = sum of roots and \(\large\frac{c}{a}\) = product of roots.
Example 2: Write a quadratic function, in standard form, with the given zeros: \(1+\large\frac{\sqrt{5}}{2}i\) and \(1-\large\frac{\sqrt{5}}{2}i\). Be sure all coefficients are integers.
\(\begin{align}&x^2 - (1+\frac{\sqrt{5}}{2}i+1-\frac{\sqrt{5}}{2}i)x+(1+\frac{\sqrt{5}}{2}i)(1-\frac{\sqrt{5}}{2}i)\ & \ \ \ &\text{1) Substitute the sum of the roots in to the quadratic equation}\\
&\ & \ \ \ &\text{for b and the product of the roots in for c.}\\
&x^2-2x+1-\frac{5}{4}i^2\ & \ \ \ &\text{2) Simplify.}\\
&x^2-2x+1+\frac{5}{4}\ & \ \ \ &\text{3) Evaluate}\ \ i^2.\\
&x^2-2x+\frac{9}{4}\ & \ \ \ &\text{4) Simplify.}\\
&y=4x^2-8x+9\ & \ \ \ &\text{5) To get integers, distribute 4 throughout the equation.}\end{align}\)
Guided Learning