58.7k views
22 votes
Find the value for k for which y= 2x+k is a tangent to the curve y=2x^2-3 and determine the point of intersection

User Feradz
by
8.3k points

1 Answer

1 vote

Answers:

k = -3.5

Intersection point is (0.5, -2.5)

================================================

Step-by-step explanation:

Apply the derivative to y=2x^2-3 and you should get dy/dx = 4x

The derivative helps determine the slope of the tangent at any point on the curve.

The slope of the tangent line y = 2x+k is 2.

We want the slope of the tangent to be 2, so we'll replace the dy/dx with 2 and solve for x.

dy/dx = 4x

2 = 4x

x = 2/4

x = 0.5

Plug this into the curve's original equation.

y = 2x^2 - 3

y = 2(0.5)^2 - 3

y = -2.5

Therefore, the tangent line y = 2x+k and the curve y = 2x^2-3 intersect at the point (0.5, -2.5). This is the point of tangency.

We'll use the coordinates of this point to determine k.

y = 2x+k

-2.5 = 2(0.5) + k

-2.5 = 1 + k

k = -2.5-1

k = -3.5

Visual verification is shown below. I used GeoGebra to make the graph, but you could use any other tool you prefer (such as Desmos).

Find the value for k for which y= 2x+k is a tangent to the curve y=2x^2-3 and determine-example-1
User Daniel Qiu
by
7.7k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories