# How do you write a quadratic function in standard form whose graph passes through points (1,2), (-2,5), (3,-20)?

Jun 12, 2017

Use the standard form:

$y = a {x}^{2} + b x + c \text{ }$

and the 3 given points to create rows in an augmented matrix.
Solve the matrix for $a , b , \mathmr{and} c$
Substitute the values into equation 

#### Explanation:

Use the point $\left(1 , 2\right)$in equation :

$a \left({1}^{2}\right) + b \left(1\right) + c = 2$

$a + b + c = 2$

This gives us the first row:

[ (1,1,1,|,2) ]

Use the point $\left(- 2 , 5\right)$ in equation :

$a \left(- {2}^{2}\right) + b \left(- 2\right) + c = 5$

$4 a - 2 b + c = 5$

This gives us the second row:

[ (1,1,1,|,2), (4,-2,1,|,5) ]

Use the point $\left(3 , - 20\right)$ in equation :

$a \left({3}^{2}\right) + b \left(3\right) + c = - 20$

$9 a + 3 b + c = - 20$

This gives us the third row:

[ (1,1,1,|,2), (4,-2,1,|,5), (9,3,1,|,-20) ]

Perform elementary two operations until we obtain an identity matrix on the left.

${R}_{2} - 4 {R}_{1} \to {R}_{2}$:

[ (1,1,1,|,2), (0,-6,-3,|,-3), (9,3,1,|,-20) ]

${R}_{3} - 9 {R}_{1} \to {R}_{3}$:

[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,-6,-8,|,-38) ]

${R}_{3} - {R}_{2} \to {R}_{3}$:

[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,0,-5,|,-35) ]

$- \frac{1}{5} {R}_{3} \to {R}_{3}$:

[ (1,1,1,|,2), (0,-6,-3,|,-3), (0,0,1,|,7) ]

${R}_{2} + 3 {R}_{3} \to {R}_{2}$:

[ (1,1,1,|,2), (0,-6,0,|,18), (0,0,1,|,7) ]

$- \frac{1}{6} {R}_{2} \to {R}_{2}$:

[ (1,1,1,|,2), (0,1,0,|,-3), (0,0,1,|,7) ]

${R}_{1} - {R}_{3} \to {R}_{1}$

[ (1,1,0,|,-5), (0,1,0,|,-3), (0,0,1,|,7) ]

${R}_{1} - {R}_{2} \to {R}_{1}$

[ (1,0,0,|,-2), (0,1,0,|,-3), (0,0,1,|,7) ]

We have an identity matrix on the left, therefore the solution is in the column vector on the right:

$a = - 2 , b = - 3 \mathmr{and} c = 7$

Substitute these values into equation :

$y = - 2 {x}^{2} - 3 x + 7 \text{ }$

To prove that this is the correct solution, here is a graph of the 3 points and equation : 