# How do you identify the important parts of f(x)=-x^2+4x-6 to graph it?

Oct 9, 2015

Graph $f \left(x\right) = - {x}^{2} + 4 x - 6$

#### Explanation:

The important parts are:
a. x-coordinate of axis of symmetry:
$x = \left(- \frac{b}{2 a}\right) = - \frac{4}{-} 2 = 2$
b x-coordinate of vertex: same as the one of axis of symmetry:
x = 2
y-coordinate of vertex:
y = f(2) = -4 + 8 - 6 = -2.
c. y-intersect. Make x = 0 --> y = -6.
d. x- intercepts. make y = 0 --> Solve: $- {x}^{2} + 4 x - 6 = 0.$
$D = {b}^{2} - 4 a c = 16 - 24 = - 8$. There are no real roots (No x-intercepts). The parabola open downward (a < 0) and it is completely below the x-axis.
graph{-x^2 + 4x - 6 [-10, 10, -5, 5]}