# What is the axis of symmetry and vertex for the graph y= -4x^2 + 3 ?

Oct 17, 2016

See explanation

#### Explanation:

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

The y-axis intercept is the constant c which in this case gives $y = 3$

As the $b x$ term is not 0 (not there) then the graph is symmetrical about the y-axis. Consequently the vertex is actually on the y-axis.

$\textcolor{b l u e}{\text{Axis of symmetry is: } x = 0}$
color(blue)("Vertex "->(x,y)=(0,3)

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$\textcolor{b r o w n}{\text{Foot Note:}}$

As the $a {x}^{2}$ term is negative the graph form is $\cap$

If the $a {x}^{2}$ term had been positive then in that instance the graph form would be $\cup$

As a general rule the axis of symmetry is at $x = \left(- \frac{1}{2}\right) \times \frac{b}{a}$

Consider the example of $y = a {x}^{2} + b x + c \text{ "->" } y = - 2 {x}^{2} + 3 x - 4$

In this case the axis of symmetry will be at:

$x = \left(- \frac{1}{2}\right) \times \frac{b}{a} \text{ "->" "(-1/2)xx3/(-2)" " =" } \frac{3}{4}$