 What is the greater of the two -intercepts of the parabola with the equation ? Tristan says:

Wouldn’t another solution (and debatably an easier one) be to just solve for x? Subtract both sides by 4, divide by -1, square root both sides, and get +2=x – 7 and -2 = x – 7. Therefore x=5 and x=9? Mike McClenathan says:

Sure, that works too! infoanalysis says:

I dont get that “If you know the normal shape of a parabola with a leading coefficient of 1 or –1, then you know it will travel 2 units left or right as it travels 4 units down to the x-axis. ” I only know -1 coefficient will be down sloping and +1 will be up sloping. Dean Kremer says:

This may be too late to help the OP, but it may help someone else. What Mike means is that you can think about the standard equation and graph of y = x^2 (with an x coefficient of 1) and the (x , y) points produced by it: (1 , 1) , (2 , 4), (3 , 9), etc. as well as the corresponding negative x-values (-1, 1), (-2 , 4), (-3 , 9). These points indicate that, when we travel two left or right, we’ll travel 4 up. If the parabola opens down because of a negative coefficient, we’d travel 4 down instead. That description will characterize *any* quadratic with a coefficient of 1; you’ll just start at a different vertex than
(0 , 0) as you would for y = x^2.
Therefore, you can start at the (h , k) vertex indicated by this equation (7 , 4), knowing you need to travel 4 down to get to the x-axis to reach the x-intercepts the question asks about. Since we’re traveling 4 down, it follows we’d have to travel 2 left and 2 right from 7, so the x-int’s are (5 , 0) and (9, 0) (which you can double-check by putting those points into the equation). The greater of these is 9.