1. ## finding x-intercepts

how do i find the x-intercept of a quadratic equation?
EX. y=5x^2

EX.y=2x^2+x+7
i also would like to know how to graph inequalities.

Ryan

2. Originally Posted by ryan southwick
how do i find the x-intercept of a quadratic equation?
EX. y=5x^2

EX.y=2x^2+x+7
i also would like to know how to graph inequalities.

Ryan
THere are three general cases of the quadratic equation $\displaystyle ax^2+bx+c=0$

Case 1 all variables are present

In this case you can attempt to factor or utilize the quadratic formula

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

This in its self leaves three possiblity

if $\displaystyle b^2-4ac>0$ There are two real solutions

if $\displaystyle b^2-4ac=0$ there is one real solution

if $\displaystyle b^2-4ac<0$ there are no real solutions this arises from the domain conflict of $\displaystyle \sqrt{b^2-4ac}$ when this is negative

Case 2 c=0

This yields $\displaystyle ax^2+bx=0$

Factoring we get $\displaystyle x(ax+b)=0$

Expanding upon that utilizing the zero-product property we get

$\displaystyle x=0$ or $\displaystyle ax+b=0\Rightarrow{x=\frac{-b}{a}}$

Case 3 b=0

This gives $\displaystyle ax^2+c=0$

solving for x we get

$\displaystyle ax^2=-c\Rightarrow{x^2=\frac{-c}{a}}\Rightarrow{x=\pm\sqrt{\frac{-c}{a}}}$

Realizing that we must stipulate that $\displaystyle c<0$ to have a real solution.

This should provide all the information on quadratics you will need

As for graphing inequalities look here Graphing Linear Inequalities

3. If you solve a quadratic equation, like the 2nd one you have above, using whatever method, you are solving it to get values of x. These values will be where the graph cuts the x axis, ie where y=o, or will show you that the graph will not cut the x-axis if real values do not exist. If a equation will factorise, you will have two separate roots, or one repeated root (ie the same value of x twice). If you have two roots, the graph will cut the x axis at these two points. If you have a repeated root, the bottom or tip of the graph will touch the x-axis at this point.
So y=2x^2+x+7 must first be put = to 0.
0=2x^2+x+7
This will not factorise simply, so using the quadratic formula as written above: a=2 b=1 c=7
-1 (plus or minus) sqrt 1^2-4x2x7 all divided by 2x2
This gives -1(plus or minus) sqrt -53 all divided by 4
As the number being sqrtd is a minus (ie b^2-4ac<0), there are no real roots for this equation, so the graph will not cut the x-axis.
For the first equation, if you consider an equation of a line y=mx+c, the c value gives where the line crosses the y axis. If you had y=x^2, putting in y=0, then x also=0, so the graph has only one root, which is 0. From your equation, y=5x^2, what has been altered is not the position of the graph, it is the shape. Replacing x^2 by 5x^2 means that the graph has been stretched along the y-axis, ie gets thinner.
Hope this makes sense!