Answer:
You would end up at (9,6)
Answer:
We can have two cases.
A quadratic function where the leading coefficient is larger than zero, in this case the arms of the graph will open up, and it will continue forever, so the maximum in this case is infinite.
A quadratic function where the leading coefficient is negative. In this case the arms of the graph will open down, then the maximum of the quadratic function coincides with the vertex of the function.
Where for a generic function:
y(x) = a*x^2 + b*x + c
The vertex is at:
x = -a/2b
and the maximum value is:
y(-a/2b)
The first three statements are correct while the last statement is incorrect.
Answer:
The answer for the first equation is,
X = 6, -4
The answer for second equation is,
X = 9,2
Answer:
1. x²-2x+15
This can't be factored, I've tried multiple ways now, but it just can't be factored.
2. y²-19y+90
Let's factor y^2−19y+90
y^2−19y+90
The middle number is -19 and the last number is 90.
Factoring means we want something like
(y+_)(y+_)
Which numbers go in the blanks?
We need two numbers that...
Add together to get -19
Multiply together to get 90
Can you think of the two numbers?
Try -9 and -10:
-9+-10 = -19
-9*-10 = 90
Fill in the blanks in
(y+_)(y+_)
with -9 and -10 to get...
(y-9)(y-10)
Answer:
(y−9) (y−10)
