The graph of a quadratic function has a vertex of (3,6) and passes through (1, 42). We can model the graph by using a function in the form of y = ax2 + bx + c, where "a," ";"
and "c" represent real numbers.
Find the value of "a".

Explanation:

Vertex form in general is

y = a(x-h)^2 + k

The given vertex is at (3,6) meaning (h, k) = (3, 6)

So h = 3 and k = 6 are plugged into the first equation shown above

y = a(x-h)^2 + k

y = a(x-3)^2 + 6

Now plug in x = 1 and y = 42, which is from (x,y) = (1,42).

So we get,

y = a(x-3)^2 + 6

42 = a(1-3)^2 + 6

At this point, we have one variable and we can isolate it like so:

42 = a(1-3)^2 + 6

42 = a(-2)^2 + 6

42 = a(4)

42 = 4a+6

4a+6 = 42

4a = 42-6

4a = 36

a = 36/4

a = 9 is the final answer

Extra info (optional section):

This means the y = a(x-h)^2 + k becomes y = 9(x-3)^2 + 6

Let's expand that out and get it into standard form

y = 9(x-3)^2 + 6

y = 9(x^2-6x+9) + 6

y = 9x^2-54x+81+6

y = 9x^2 - 54x + 87

The last equation is in the form y = ax^2+bx+c where,

a = 9

b = -54

c = 87

what does muffins have to do with anything?

step-by-step explanation:

see below

step-by-step explanation:

x+2> 6

subtract 2 from each side

x+2-2 > 6-2

x > 4

there is an open circle at 4 since is it is greater than and the line goes to the right