We know the y intercept is -6 and we know the slope is positive 4 (rise over run) so the equation is y=4x-6 if we plug in a shaded point (I would choose 0,0 for convenience reasons) 0=-6 since -6 is less than 0, the expression would be y≥4x-6
(8,1) since the midpoint is 3 away from the endpoint, it must also be 3 away from the other endpoint
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)
