The quadratic equation \((-x^2+2x+4)\) have the coefficient of \(x^2\) less than zero therefore its graph has a maximum point and it also have y-intercept equal to 4.
The quadratic equation that have the coefficient of \(x^2\) less than zero, then the graph of that quadratic equation have a maximum point.
Therefore according to the given equation the quadratic equation that have a maximum point are: \((-4x^2+8x +5)\;and\;(-x^2+2x+4)\).
Now, to evaluate the value of y-intercept, put the value of x equal to zero.
a). \(4x^2+6x-1\)
put x = 0 in above equation to get the y-intercept:
\(4(0)^2+6(0)-1=-1\)
So, the equation \(4x^2+6x-1\) has y-intercept -1.
b). \(-4x^2+8x+5\)
put x = 0 in above equation to get the y-intercept:
\(-4(0)^2+8(0)+5=5\)
So, the equation \(-4x^2+8x+5\) has y-intercept 5.
c). \(-x^2+2x+4\)
put x = 0 in above equation to get the y-intercept:
\(-(0)^2+2(0)+4=4\)
So, the equation \(-x^2+2x+4\) has y-intercept 4.
d). \(x^2+4x-4\)
put x = 0 in above equation to get the y-intercept:
\((0)^2+4(0)-4=-4\)
So, the equation \(x^2+4x-4\) has y-intercept -4.
Therefore, the correct option is c).
For more information, refer the link given below:
https://brainly.com/question/11243794
