How od I work out the x intercept and vertex in these two equations?

y=x^2-x-6

and

y=-x^2+3x+10

Is the x intercept purely the coefficient of x?

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- Nov 2nd 2006, 11:25 AMYppolitiaDececting Intercepts and Vertexs in Equations
How od I work out the x intercept and vertex in these two equations?

y=x^2-x-6

and

y=-x^2+3x+10

Is the x intercept purely the coefficient of x? - Nov 2nd 2006, 11:52 AMtopsquark
You know what a y intercept is, yes? The point on a graph where the function crosses the y axis. So an x intercept is a point on a graph where the function crosses the x axis.

So to find the x intercepts of $\displaystyle y = x^2 - x - 6$ you need to solve $\displaystyle x^2 - x - 6 = 0$ for x. (Note: A function can only have 1 y intercept, but it can have many x intercepts.) I should add the the x and y intercepts are*points*, so you should lable your x intercepts in the form (x, 0). (Some people get sloppy and just give an x value, which technically isn't correct.)

To find the vertex point, note that the axis of symmetry of the parabola $\displaystyle y = ax^2 + bx + c$ is the line $\displaystyle x = -\frac{b}{2a}$. So find x = -b/(2a) for your parabola and plug that x value into the function to find y. Then the pair (x, y) is your vertex point.

-Dan