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Find at the point (x,y), at which the graph of y=2x² 3x-7 has a horizontal tangent

User Jazz
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Final answer:

To find the point of the graph with a horizontal tangent of the equation y = 2x² + 3x - 7, we need to find the derivative and set it equal to zero. The point is (-3/4, -37/8).

Step-by-step explanation:

To find the point at which the graph of the equation y = 2x² + 3x - 7 has a horizontal tangent, we need to find the derivative of the equation and set it equal to zero.

The derivative of y = 2x² + 3x - 7 is dy/dx = 4x + 3. Setting dy/dx = 0, we have 4x + 3 = 0. Solving for x, we find x = -3/4. Substituting this value of x back into the original equation, we can find the y-coordinate.

When x = -3/4, plugging in this value into the original equation y = 2x² + 3x - 7, we get y = 2(-3/4)² + 3(-3/4) - 7 = -37/8. Therefore, the point at which the graph has a horizontal tangent is (-3/4, -37/8).

User Anter
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