(QUICK) The vertex of this parabola is at (2, -1). When the y-value is 0, the x-value is 5. What is the coefficient of the squared term in the parabola's equation? (QUICK) The vertex of this parabola is at (2, -1). When the y-value is 0, the x-value is 5. What is the coefficient of the squared term in the parabola's equation? https://gyazo.com/359a21c5c3450dd4b7aec64017e5246e

Question

(QUICK) The vertex of this parabola is at (2, -1). When the y-value is 0, the x-value is 5. What is the coefficient of the squared term in the parabola's equation? (QUICK) The vertex of this parabola is at (2, -1). When the y-value is 0, the x-value is 5. What is the coefficient of the squared term in the parabola's equation?  https://gyazo.com/359a21c5c3450dd4b7aec64017e5246e

campbell_st · Answer

you have h  2 and k = -1

substitute them into 

$$y = a(x -h)^2 + k$$

next use the point (5, 0) 

substitute x = 5 and y = 0

and solve for a

thecrazy_t · Answer

so far 0=a(5-2)^2+-1 correct? then solve that as normal for a

thecrazy_t · Answer

is not one of my options though

campbell_st · Answer

you need to calculate the answer so 

0 = 9a - 1

now solve for a, that will be the coefficient of x^2