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Math Help - Dececting Intercepts and Vertexs in Equations

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    Dececting 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?
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    Quote Originally Posted by Yppolitia View Post
    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?
    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 y = x^2 - x - 6 you need to solve 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 y = ax^2 + bx + c is the line 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
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