How many times does the graph of the function below interect or touch the x-axis? y=-2x^2+3x+5

Two times at (-1,0) and (2.5,0)

Step-by-step explanation:

When the graph intersects or touches x-axis, y is equal to 0

so y = -2x^2 + 3x + 5

=> 0 = -2x^2 + 3x + 5

The formula to solve a quadratic equation of the form ax^2 + bx + c = 0 is equal to x = [-b +/-√(b^2 - 4ac)]/2a

so a = -2

    b = 3

    c = 5

substitute in the formula

x = [-3 +/- √(3^2 - 4x-2x5)]/2(-2)

x = [-3 +/- √(9 + 40)]/(-4)

x = [-3 +/- 7]/(-4)

x1 = (-3 + 7)/(-4) = 4/-4 = -1

x2 = (-3 - 7)/(-4) = -10/-4 = 5/2 = 2.5

so the graph has two x-intercepts (-1,0) and (2.5,0), therefore it intersects x-axis two times

answer: evaluate the objective function at each vertex to determine which - and -values, if any, maximize or minimize the function. example: find the minimum value and maximum value of the objective function f ( x , y ) = 4 x + 5 y , subject to the following constraints