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?

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.

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.

## Comments (4)

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?

Sure, that works too!

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.

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.