It’s interesting that the student understand the coefficients in question determine both the dilation and whether it is reflected in the x-axis, but doesn’t seem to think that one coefficient could handle both of those things. Like, we need one 2 to make sure it points upward, and another 2 to make sure it gets dilated. Also, the fact that there is a 2 attached to the x inside the parentheses makes me think that the student is trying to cope with this as a horizontal dilation rather than as a vertical contraction since it (a) is attached to the x, and (b) has absolute value >1. In fact, maybe that’s why the student is throwing in two coefficients: the first 2 ensures that the parabola points upward (a y-value concern), and the second 2 doubles the x-coordinate (a horizontal stretching concern). The second 2 makes more sense to me. The first one? I would ask why a positive 1 wouldn’t suffice there to make sure it points up, because the kid doesn’t think it’s dilated in both directions…at least it doesn’t seem that way.

The student understands that the vertex location indicates what h and k should be and uses the k value correctly. However, the h value is not used correctly. The student understands that the parabola must have a coefficient for the x that is not 1, and that a 2 is somehow involved, however doesn’t use the correct value. The student understands that the leading coefficient will be positive, however for some reason uses 2.

I would first have the student look at the structure of the graph and see that the roots are easy to see. Therefore, the easiest approach would be to use the y = a(x-1)(x+3) format for approaching the equation. The “a” will need to be solved for by plugging in another known point that is not a root. There are several to choose from. The final answer would be y = .5(x-1)(x+3), which is a valid form to leave it in. However, if the student wanted to find the h,k form, then completing the square with all of the requisite algebra would be needed.

## 2 replies on “Equation of Parabola”

It’s interesting that the student understand the coefficients in question determine both the dilation and whether it is reflected in the x-axis, but doesn’t seem to think that

onecoefficient could handle both of those things. Like, we needone2 to make sure it points upward, andanother2 to make sure it gets dilated. Also, the fact that there is a 2 attached to the x inside the parentheses makes me think that the student is trying to cope with this as a horizontal dilation rather than as a vertical contraction since it (a) is attached to the x, and (b) has absolute value >1. In fact, maybe that’s why the student is throwing in two coefficients: the first 2 ensures that the parabola points upward (a y-value concern), and the second 2 doubles the x-coordinate (a horizontal stretching concern). The second 2 makes more sense to me. The first one? I would ask why a positive 1 wouldn’t suffice there to make sure it points up, because the kid doesn’t think it’s dilated in both directions…at least it doesn’t seem that way.The student understands that the vertex location indicates what h and k should be and uses the k value correctly. However, the h value is not used correctly. The student understands that the parabola must have a coefficient for the x that is not 1, and that a 2 is somehow involved, however doesn’t use the correct value. The student understands that the leading coefficient will be positive, however for some reason uses 2.

I would first have the student look at the structure of the graph and see that the roots are easy to see. Therefore, the easiest approach would be to use the y = a(x-1)(x+3) format for approaching the equation. The “a” will need to be solved for by plugging in another known point that is not a root. There are several to choose from. The final answer would be y = .5(x-1)(x+3), which is a valid form to leave it in. However, if the student wanted to find the h,k form, then completing the square with all of the requisite algebra would be needed.