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# how do graph quadratic equations?

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Jun 11, 2014

#2
+102466
+5

$$y=ax^2+bx+c\qquad a\ne0$$

this is a parabola.

If a>0 then it is concave up, If a<0 then it is concave down.

The y intercept is c

$$\mbox{The axis of symmetry is }\qquad x=\dfrac{-b}{2a}\\ \mbox{Substitute in this value of x to get the y co-ordinate of the vertex}$$

The roots, x intercepts are

$$x=\dfrac{-b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}$$

That is usually enough.

Jun 12, 2014

#1
+259
0

Please, put your question in a more appriate and mathematically correct shape. Your question is too general.

Jun 11, 2014
#2
+102466
+5

$$y=ax^2+bx+c\qquad a\ne0$$

this is a parabola.

If a>0 then it is concave up, If a<0 then it is concave down.

The y intercept is c

$$\mbox{The axis of symmetry is }\qquad x=\dfrac{-b}{2a}\\ \mbox{Substitute in this value of x to get the y co-ordinate of the vertex}$$

The roots, x intercepts are

$$x=\dfrac{-b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}$$

That is usually enough.

Melody Jun 12, 2014