Final answer:
To find the horizontal tangent of the function y = 2x^2 + 3x - 8, the derivative is calculated and set to zero to find x, then y is found by plugging x back into the original equation. However, the result does not match any of the provided options, suggesting an error in the question or options.
Step-by-step explanation:
To find the point (x, y) at which the graph of y = 2x^2 + 3x - 8 has a horizontal tangent, we need to determine where the derivative of the function is equal to zero. The derivative represents the slope of the tangent to the curve at any point.
The derivative of y with respect to x is given by:
dy/dx = 4x + 3.
For a horizontal tangent, the slope must be zero:
0 = 4x + 3
Solving this for x gives us x = -3/4.
We then substitute x back into the original equation to find y:
y = 2(-3/4)^2 + 3(-3/4) - 8 = -6.5.
There is a small issue, however, as none of the options given (A) (1, 1) B) (2, -1) C) (-2, -5) D) (-1, 4)) match the coordinates we calculated. This could indicate a possible mistake in the question or the provided options.