# What is the equation of the parabola with axis intercepts of x=-6, x=5, and y=3?

May 31, 2016

It is $y = - \frac{1}{10} {x}^{2} - \frac{1}{10} x + 3$.

#### Explanation:

The parabola has equation

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

and we have to find three parameters to determine it: $a , b , c$.
To find them we have to use the three given points that are
$\left(- 6 , 0\right) , \left(5 , 0\right) , \left(0 , 3\right)$. The zeros are because the points are intercepts, it means that in those points they cross or the $y$ axes (for the first two) or the $x$ axes (for the last one).
We can substitute the values of the points in the equation

$0 = a \cdot {\left(- 6\right)}^{2} + b \cdot \left(- 6\right) + c$
$0 = a \cdot {5}^{2} + b \cdot 5 + c$
$3 = a \cdot {0}^{2} + b \cdot 0 + c$

I do the calculations and have

$0 = 36 a - 6 b + c$
$0 = 25 a + 5 b + c$
$3 = c$

We are lucky! From the third equation we have the value of $c$ that we can use in the first two, so we have

$0 = 36 a - 6 b + 3$
$0 = 25 a + 5 b + 3$
$3 = c$

We find $a$ from the first equation

$0 = 36 a - 6 b + 3$
$36 a = 6 b - 3$
$a = \frac{6 b - 3}{36} = \frac{b}{6} - \frac{1}{12}$

and we substitute this value in the second equation

$0 = 25 a + 5 b + 3$
$0 = 25 \left(\frac{b}{6} - \frac{1}{12}\right) + 5 b + 3$
$0 = \frac{25}{6} b + 5 b + 3 - \frac{25}{12}$
$0 = \frac{25 + 30}{6} b + \frac{36 - 25}{12}$
$0 = \frac{55}{6} b + \frac{11}{12}$
$\frac{55}{6} b = - \frac{11}{12}$
$b = - \frac{1}{10}$.

And finally I use this value of $b$ in the previous equation

$a = \frac{b}{6} - \frac{1}{12}$
$a = - \frac{1}{10} \cdot \frac{1}{6} - \frac{1}{12} = - \frac{1}{60} - \frac{1}{12} = - \frac{1}{60} - \frac{5}{60} = - \frac{6}{60} = - \frac{1}{10}$

Our three numbers are $a = - \frac{1}{10} , b = - \frac{1}{10} , c = 3$ and the parabola is

$y = - \frac{1}{10} {x}^{2} - \frac{1}{10} x + 3$. We can verify looking if the plot pass for the three points $\left(- 6 , 0\right) , \left(5 , 0\right) , \left(0 , 3\right)$.

graph{y=-1/10x^2-1/10x+3 [-10, 10, -5, 5]}