find the point $(x,y)$ at which the graph of $y = 5x^{2}+4x - 2$ has a horizontal tangent line.\nthe…

find the point $(x,y)$ at which the graph of $y = 5x^{2}+4x - 2$ has a horizontal tangent line.\nthe function $y = 5x^{2}+4x - 2$ has a horizontal tangent line at $square$.\n(type an ordered pair. simplify your answer.)

find the point $(x,y)$ at which the graph of $y = 5x^{2}+4x - 2$ has a horizontal tangent line.\nthe function $y = 5x^{2}+4x - 2$ has a horizontal tangent line at $square$.\n(type an ordered pair. simplify your answer.)

Answer

Explanation:

Step1: Find the derivative of the function

The derivative of (y = 5x^{2}+4x - 2) using the power rule ((x^{n})^\prime=nx^{n - 1}) is (y^\prime=\frac{d}{dx}(5x^{2}+4x - 2)=10x + 4).

Step2: Set the derivative equal to zero

Since the slope of a horizontal tangent line is (0), we set (y^\prime = 0). So, (10x+4 = 0). Solve for (x): (10x=-4), then (x =-\frac{4}{10}=-\frac{2}{5}).

Step3: Find the (y)-coordinate

Substitute (x =-\frac{2}{5}) into the original function (y = 5x^{2}+4x - 2). (y=5\times\left(-\frac{2}{5}\right)^{2}+4\times\left(-\frac{2}{5}\right)-2) (y = 5\times\frac{4}{25}-\frac{8}{5}-2) (y=\frac{4}{5}-\frac{8}{5}-2) (y=\frac{4 - 8}{5}-2) (y=-\frac{4}{5}-2=-\frac{4 + 10}{5}=-\frac{14}{5})

Answer:

(\left(-\frac{2}{5},-\frac{14}{5}\right))