I know how to sketch a graph when two real roots, how do I do it with 1 or equal?

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    I know how to sketch a graph when there are two real roots, and I have attached a solution that is the way I Iearnt to do it, by substituting values in, factorising/solving for x coordinates, then substituting x into the equation to find the y coordinates.

    However, how would I do it if there were A) 1 real root, and B) 0 real roots? I know that if there are no real roots, then b^2-4ac is equal to 0. I don't know how you would go about solving (and doing what I do in this example) when there are no real roots though. Same goes for if there is 1 real root, but I don't have a clue how you would solve for x and y crossover coordinate points, and I don't know how you would do something like this solution when there is 1 real root.
    Please could somebody explain how you do like what I do in this solution (when there are two real roots), but when there are A) 0 real roots and B) 1 real root?? The book has no other examples.
    Thanks, it would be greatly appreciated.

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    If there are repeated roots (1 root) the local minimum lies on the x axis. If there are no real roots, the curve is above the x axis.
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    (Original post by blobbybill)
    ...
    For 0 real roots, the parabola is strictly above the x-axis. Discriminant is b^2-4ac<0

    For 1 real root (or repeated roots, in other words), the vertex of the parabola touches the x-axis (ie is tangent to it). Discriminant is b^2-4ac=0

    For 2 real roots, the parabola strictly intersects the x-axis at 2 different points as in the example above. Discriminant is b^2-4ac>0

    Everything else is just the matter of labelling, such as the y-intercept which is just the constant of the equation (or when x=0).
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    (Original post by RDKGames)
    For 0 real roots, the parabola is strictly above the x-axis. Discriminant is b^2-4ac<0

    For 1 real root (or repeated roots, in other words), the vertex of the parabola touches the x-axis (ie is tangent to it). Discriminant is b^2-4ac=0

    For 2 real roots, the parabola strictly intersects the x-axis at 2 different points as in the example above. Discriminant is b^2-4ac>0

    Everything else is just the matter of labelling, such as the y-intercept which is just the constant of the equation (or when x=0).
    How would I use something like the solution I attached in order to find out where the x coordinate and y coordinate is on one with 1 real root? And how would I use that solution method to work out the y coordinate when there are no real roots?
    Is it just the same as the solution I attached, but you miss a few steps because there is only 1 x crossover point coordinate (with 1 real root), and no x crossover point (with no real roots)?
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    (Original post by blobbybill)
    How would I use something like the solution I attached in order to find out where the x coordinate and y coordinate is on one with 1 real root? And how would I use that solution method to work out the y coordinate when there are no real roots?
    Is it just the same as the solution I attached, but you miss a few steps because there is only 1 x crossover point coordinate (with 1 real root), and no x crossover point (with no real roots)?
    What are you actually asking? I don't understand.

    The method is always the same. The discriminant tells you how many real roots there are, then you solve the quadratic to get the roots, then you plug in x=0 to see where the parabola crosses the y-axis, note those points down on a graph, and sketch the parabola. If you want accuracy then you'd need to find the vertex as well which is a simple case of completing the square.
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    (Original post by RDKGames)
    What are you actually asking? I don't understand.

    The method is always the same. The discriminant tells you how many real roots there are, then you solve the quadratic to get the roots, then you plug in x=0 to see where the parabola crosses the y-axis, note those points down on a graph, and sketch the parabola. If you want accuracy then you'd need to find the vertex as well which is a simple case of completing the square.
    That's what I wanted to know. I was wondering if you always found the number of real roots, then solved quadratic, plugged in x=0, etc. Thanks.

    What do you mean by "if you want accuracy"? What is a vertex?
    Sorry for my bad explanation of my question.
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    (Original post by blobbybill)
    That's what I wanted to know. I was wondering if you always found the number of real roots, then solved quadratic, plugged in x=0, etc. Thanks.

    What do you mean by "if you want accuracy"? What is a vertex?
    Sorry for my bad explanation of my question.
    The vertex is the point on the curve through which the line of symmetry for the quadratic goes through.
 
 
 
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