suppose that f(x)=5x² - 4x. find: (a) f(x)=10x - 4 (b) the slope of the graph of f(x) at x = 2 and x = 3…

Explanation:

Step1: Recall point - slope form

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a point on the line. First, find $y$ - values for $x = 2$ and $x = 3$ in $f(x)=5x^{2}-4x$.

Step2: Calculate $y$ - value at $x = 2$

Substitute $x = 2$ into $f(x)$: $f(2)=5\times2^{2}-4\times2=5\times4 - 8=20 - 8 = 12$. The slope $m_1$ at $x = 2$ is $f^{\prime}(2)=16$. Using the point - slope form $y - 12=16(x - 2)$, expand to get $y=16x-32 + 12=16x-20$.

Step3: Calculate $y$ - value at $x = 3$

Substitute $x = 3$ into $f(x)$: $f(3)=5\times3^{2}-4\times3=5\times9-12 = 45 - 12=33$. The slope $m_2$ at $x = 3$ is $f^{\prime}(3)=26$. Using the point - slope form $y - 33=26(x - 3)$, expand to get $y=26x-78 + 33=26x-45$.

Step4: Find when tangent is horizontal

A horizontal tangent has slope $m = 0$. Set $f^{\prime}(x)=0$. Since $f^{\prime}(x)=10x - 4$, then $10x-4 = 0$. Solve for $x$: $10x=4$, so $x=\frac{2}{5}$.

Answer:

Tangent line at $x = 2$: $y = 16x-20$ Tangent line at $x = 3$: $y = 26x-45$ Value(s) of $x=\frac{2}{5}$