How do you write a quadratic equation with vertex of (3,-6) and the point (-1,10) ?

Mar 27, 2017

I have taken you to a point where you can finish it off.

Explanation:

Consider the standardised form of $y = a {x}^{2} + b x + c$

Then ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times b$

$\implies$ given that vertex$\to \left(x , y\right) = \left(3 - 6\right)$ then we have

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times b = 3$

$\implies \textcolor{m a \ge n t a}{b = 3 \times \left(- \frac{2}{1}\right) = - 6}$

So we now have:

$y = a {x}^{2} - 6 x + c$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
2 unknowns and 2 sets of values so solvable.

For: $\left(x , y\right) \to \left(3 , - 6\right)$
$\implies - 6 = a \left({3}^{2}\right) - 6 \left(3\right) + c \text{ } \ldots E q u a t i o n \left(1\right)$

For: $\left(x , y\right) \to \left(- 1 , 10\right)$
$\implies 10 = a {\left(- 1\right)}^{2} - 6 \left(- 1\right) + c . E q u a t i o n \left(2\right)$

$- 6 = 9 a - 18 + c \text{ } \ldots E q u a t i o n \left(1\right)$
$\textcolor{w h i t e}{.} 10 = \textcolor{w h i t e}{.} a + \textcolor{w h i t e}{.} 6 + c \text{ } . . E q u a t i o n \left(2\right)$

Multiply $E q u a t i o n \left(2\right)$ by 9

$\textcolor{w h i t e}{.} 90 = 9 a + 54 + 9 c \text{ } . . E q u a t i o n \left({2}_{a}\right)$
ul(-6=9a-18+c" ".....Equation(1))larr" subtract"
$84 = 0 a + 72 + 10 c$

$10 c = 12$

$\textcolor{m a \ge n t a}{c = \frac{6}{5}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now that you have $c$ yo can determine the value of $a$ by substitution. I will let you do that.