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

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Answer 1 · 2018-06-08T19:24:19

See graph

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

first put in standard form:

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

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

$y=0^2+4(0)-3$

$y=-3$

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

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

factor:

$0=-(x - 1)(x - 3)$

$x=1$ and $x=3$

Axis of Symmetry (aos) in the form $ax^2+bx+c$

$aos = (-b)/(2a)$

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

$aos = (-4)/(2(-1))$

$aos=2$

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

$vertex = (aos, f(aso))$

$vertex = (2, f(2))$

$vertex = (2, -2^2+4*2-3)$

$vertex = (2, 1)$ is a maximum.

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

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