Question

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

Answer 1

Answers:  

k = -3.5

Intersection point is (0.5, -2.5)

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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).