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.
For 0 real roots, the parabola is strictly above the x-axis. Discriminant is b2−4ac<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 b2−4ac=0
For 2 real roots, the parabola strictly intersects the x-axis at 2 different points as in the example above. Discriminant is b2−4ac>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)?
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.
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.