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.

please help!!!

Ryan

Printable View

- May 9th 2008, 03:16 PMryan southwickfinding 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.

please help!!!

Ryan - May 9th 2008, 04:03 PMMathstud28
THere are three general cases of the quadratic equation

Case 1 all variables are present

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

This in its self leaves three possiblity

if There are two real solutions

if there is one real solution

if there are no real solutions this arises from the domain conflict of when this is negative

Case 2 c=0

This yields

Factoring we get

Expanding upon that utilizing the zero-product property we get

or

Case 3 b=0

This gives

solving for x we get

Realizing that we must stipulate that 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 - May 10th 2008, 02:05 AMbbbabybel
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!