The graph of a quadratic function has a vertex ar (3,-6). one point of the graph is (7,10). what is another point on the graph?

y + 6 = (x -3)^2

Step-by-step explanation:

Let's use the general form of the vertex equation of a quadratic:  y - k = a(x - h)^2.

Subst. 10 for y, 7 for x, -6 for k and 3 for h.  Then, 10 + 6 = a(7 - 3)^2, or

16 = a(4)^2, or 16 = a(16).  Thus, a = 1, and the equation of this particular parabola is:

y + 6 = (x -3)^2

0 is its x and y coordinate

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116