# How to determine the x-intercepts of a quadratic equation in standard form without a "c" value. Y=ax^2+bx

The x-intercept(s) of a quadratic in the form y=ax^2+bx are 0 and -b/a.

We are asked to determine the x-intercepts of a quadratic in the form `y=ax^2+bx` . The x-intercepts of a function y=f(x) occur wherever y=0.

The most direct way is to factor out the greatest common factor, which will include x.

`y=ax^2+bx=x(ax+b)`

The right hand side is now written as a product of two factors. Using the zero-product property (if mn=0 then either m=0, n=0, or m=n=0), we see that either x=0 or ax+b=0. Solving for x in the second equation, we subtract b then divide by a to get `x=-b/a` so `x=0 " or " x=-b/a` .

We can check these:

`x=0 => y=a(0)^2+b(0)=0`

`x=-b/a => y=a((-b)/a)^2+b((-b)/a)=a(b^2/a^2)-b^2/a=0`

For example, `y=2x^2-3x` has x-intercepts 0 and 3/2.

If `y=4x^2+6x`, then a=4, b=6, so the x-intercepts are 0 and -3/2. It is difficult to try to memorize "rules" for every possible situation, but you can factor out the common factor of 2x to get y=2x(2x+3) so 2x=0 or 2x+3=0 to get the same result.

If the quadratic is of the form `y=ax^2+bx`, then its graph is a parabola with axis of symmetry `x=(-b)/(2a)` . It is clear that zero is an x-intercept; the axis of symmetry is the perpendicular bisector of the segment joining the x-intercepts so is twice the distance from the origin or `2((-b)/(2a))=-b/a`