Question

find all points (x,y) on the graph of f(x)=2x² - 5x with tangent lines parallel to the line y = 3x + 1. the point(s) is/are. (type an ordered pair. use a comma to separate answers as needed.)

Explanation:

Step1: Find the derivative of \(f(x)\)

The derivative of \(f(x)=2x^{2}-5x\) using the power rule \((x^{n})^\prime = nx^{n - 1}\) is \(f^\prime(x)=\frac{d}{dx}(2x^{2}-5x)=2\times2x-5 = 4x-5\).

Step2: Determine the slope of the given line

The line \(y = 3x+1\) is in the form \(y=mx + b\) (slope - intercept form), where \(m\) is the slope. So, the slope of the line \(y = 3x + 1\) is \(m = 3\).

Step3: Set the derivative equal to the slope of the line

Since the tangent lines are parallel to \(y = 3x+1\), we set \(f^\prime(x)=3\).

$$
LATEXBLOCK0
$$

Step4: Find the \(y\) - value

Substitute \(x = 2\) into \(f(x)=2x^{2}-5x\).

$$
LATEXBLOCK1
$$

Answer:

\((2,-2)\)