Function: y=−x2+3x+8

Sure, I can help with that. The function you've given is a quadratic function. Its general form is y = ax^2 + bx + c, where a, b, and c are constants.

In your function, y = -x^2 + 3x + 8, the constants are a = -1, b = 3, and c = 8.

If you want to find the roots of the function (i.e., the x-values for which y = 0), you can set the function equal to zero and solve for x:

0 = -x^2 + 3x + 8

To solve this quadratic equation, you can use the quadratic formula, which is x = [-b ± sqrt(b^2 - 4ac)] / (2a).

Substituting the values of a, b, and c into the formula gives:

x = [-3 ± sqrt((3)^2 - 4*(-1)8)] / (2-1) x = [-3 ± sqrt(9 + 32)] / -2 x = [-3 ± sqrt(41)] / -2

So the roots of the function are x = [-3 + sqrt(41)] / -2 and x = [-3 - sqrt(41)] / -2.

If you want to find the vertex of the parabola, you can use the formula for the x-coordinate of the vertex, which is -b / (2a). Substituting the values of a and b gives:

x = -3 / (2*-1) = 3/2

Substituting x = 3/2 into the function gives the y-coordinate of the vertex:

y = -(3/2)^2 + 3*(3/2) + 8 = -9/4 + 9/2 + 8 = 23/4

So the vertex of the parabola is (3/2, 23/4).

The parabola opens downwards because the coefficient of x^2 is negative.

I hope this helps! Let me know if you have any other questions.