Got a test tomorrow, I'd appreciate some help. I'm pretty good in math but I just don't get these questions here.
Find, in the form y = ax2 + bx + c, the equation of the quadratic whose graph:
q1) touches the x-axis at 4 and passes through (2,12)
q2) has vertex (-4, 1) and passes through (1, 11)
You need to have 2 equations for the 2 different situations. It's the same technique but I just don't know it. Any help appreciated!!
-b/2a = 4
b = -8a
Then I replace 2 with x and y with 12?
4a + 2b + c = 12
16a + 4b + c = 0 (x intercept replaced)
12a + 2b = -12
Since b is -8a --> 12a + 2(-8a) = -12
-4a = - 12
a = 3
Afterwards I just replace 3 with a in the previous 12 + 2b = -12 equation.
b = -24
I find c the same way, it's 48.
So I guess the final equation should be y = 3x2 - 24x + 48
I still don't understand the theory though.. why is -b/2a = 4??
Isn't 4 the x-intercept, but not the minimum/maximum point of the graph?
As for the vertex, we have
Completing the square would give:
You should know that the vertex of a quadratic of the form (known as the vertex form of a quadratic) is and this is what we have here. and
If you're still unsure as to why this is the case, you should probably consult an instructor.