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)
Answer:
B,D,F
Step-by-step explanation:
ive answered harder haha
hope this helps
Answer:
y=12/5−3x/5
Step-by-step explanation:
might be wrong
To find the slope of the line:
(y2-y1)/(x2-x1)
(14-5)/(4-1)=(9)/(3)=3
Only one of your 4 possible answers has 3 as a slope. However, plugging in each point into the y=mx+b equation, the y-intercept consistently comes out as 2..
y=mx+b
14=3(4)+b b=2
5=3(1)+b b=2
y=3x+2
If there is a consistent, positive slope (from your question, this does not seem to have a quadratic as an option), 3x+5 is not even a viable solution because x=1 when the y-value is 5 (and thus no other x value {0} could have a y-value of 5). It seems as though you have a typo on your hands. Hopefully this helps?
Answer:
c and g
Step-by-step explanation:
