# How do you write the quadratic in vertex form given vertex is (1/10, -9/10) and y intercept is -1?

May 15, 2015

Vertex form:$f \left(x\right) = a {\left(x - \frac{a}{2} b\right)}^{2} + f \left(- \frac{a}{2} b\right)$
Standard form f(x) = ax^2 + bx + c
First find a, b, and c. We have as information:

x of vertex:$\left(- \frac{b}{2} a\right) = \frac{1}{10} \to a = \frac{- 10 b}{2} = - 5 b$ (1)

y intercept (-1) gives -> c = -1

y of vertex:$f \left(- \frac{b}{2} a\right) = \left(- \frac{9}{10}\right) = \frac{a}{100} + \frac{b}{10} - 1$ ->

a + 10b - 100 + 90 = 0 -> a + 10b - 10 = 0 . Replace a by -5b

-5b + 10b = 10 -> b = 10/5 = 2 -> a = -10.

Standard form: $f \left(x\right) = - 10 {x}^{2} + 2 x - 1$

Finally, vertex form: $f \left(x\right) = - 10 {\left(x - \frac{1}{10}\right)}^{2} - \frac{90}{100}$

Check: Develop f(x)
$f \left(x\right) = - 10 \left({x}^{2} - \frac{2 x}{10} + \frac{1}{100}\right) - \frac{90}{100} =$
$= - 10 {x}^{2} + 2 x - \frac{10}{100} - \frac{90}{100} = - 10 {x}^{2} + 2 x - 1$. Correct