Question

Q1.find the point on the curve y=2x^2-6x-4 at which the tangent is parallel to the x-axis

Answer 1

Step-by-step explanation:

Question : Find the point on the curve y=2x^2-6x-4 at which the tangent is parallel to the x-axis

Parallel lines have same Slope

So, Slope of Tangent = Slope of X-axis

Slope of X axis is 0

⇒ Slope of Tangent = 0

Given curve,

y = 2x² - 6x - 4

Slope of the tangent is given by,

$\begin{gathered} \frac{dy}{dx} = \frac{d}{dx} (2 {x}^{2} - 6x - 4) \\ \\ m = 4x - 6\end{gathered}$

Given Slope of the tangent m = 0

$\begin{gathered} \implies0 = 4x - 6 \\ \\ \implies4x = 6 \\ \\ \implies \: x = \frac{6}{4} = \frac{3}{2} \end{gathered}$

When x is 3/2,

$\begin{gathered}y = 2( \frac{3}{2} ) ^{2} - 6( \frac{3}{2} ) - 4 \\ \\ y = 2( \frac{9}{4}) - 9 - 4 \\ \\ y = \frac{9}{2} - 13 \\ \\ y = \frac{9 - 26}{2} \\ \\ y = \frac{ - 17}{2} \end{gathered}$

Therefore,

$(x, y) = \frac{3}{2}, \frac{ - 17}{2}$

is the point on the curve at which the tangent is parallel to the x axis.