Given the function f(x):
[tex]f(x)=3x^2-5x-4[/tex]
We will find the point on the given parabola where the slope of the tangent is horizontal
The horizontal line has a slope = zero
The slope of the line is the first derivative of the function
So, we will find the first derivative and equate it to zero
[tex]\begin{gathered} f^{\prime}(x)=0 \\ 3(2x)-5(1)-(0)=0 \end{gathered}[/tex]
solve the equation to find (x):
[tex]\begin{gathered} 6x-5=0 \\ 6x=5 \\ x=\frac{5}{6} \end{gathered}[/tex]
substitute with the value of (x) into the given function to find the y-coordinate
[tex]y=3(\frac{5}{6})^2-5(\frac{5}{6})-4=\frac{-73}{12}[/tex]
So, the answer will be the point will be as follows:
[tex](x,y)=(\frac{5}{6},\frac{-73}{12})[/tex]
See the following figure:
