# How do you find the vertex and intercepts for x^2-4x+y+3=0?

Jun 8, 2018

See graph

#### Explanation:

${x}^{2} - 4 x + y + 3 = 0$

first put in standard form:

$y = - {x}^{2} + 4 x - 3$

y-intercept, set $x = 0$ and solve for $y$

$y = {0}^{2} + 4 \left(0\right) - 3$

$y = - 3$

x-intercept(s) if they exist, set $y = 0$ and solve for $x$

$0 = - {x}^{2} + 4 x - 3$

factor:

$0 = - \left(x - 1\right) \left(x - 3\right)$

$x = 1$ and $x = 3$

Axis of Symmetry (aos) in the form $a {x}^{2} + b x + c$

$a o s = \frac{- b}{2 a}$

$y = - {x}^{2} + 4 x - 3$

$a o s = \frac{- 4}{2 \left(- 1\right)}$

$a o s = 2$

Finally find the vertex it is a minimum if $a > 0$ a maximum if $a < 0$

$v e r t e x = \left(a o s , f \left(a s o\right)\right)$

$v e r t e x = \left(2 , f \left(2\right)\right)$

$v e r t e x = \left(2 , - {2}^{2} + 4 \cdot 2 - 3\right)$

$v e r t e x = \left(2 , 1\right)$ is a maximum.

graph{-x^2+4x-3 [-7.21, 12.79, -8.2, 1.8]}