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

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.

Screenshot_8.png
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.
Original post by blobbybill
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For 0 real roots, the parabola is strictly above the x-axis. Discriminant is b24ac<0b^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 b24ac=0b^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 b24ac>0b^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).
(edited 7 years ago)
Reply 3
Original post by RDKGames
For 0 real roots, the parabola is strictly above the x-axis. Discriminant is b24ac<0b^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 b24ac=0b^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 b24ac>0b^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)?
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.
Reply 5
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.
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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