Question

Find k such that the line is tangent to the graph of the function. Function: f(x)=k-x²

Line: y=-6x+1

Answer 1

k = -8

Explanation:

Given both functions, y = k - x² and y = - 6x + 1. A line tangent to the graph means that the line has one common point with the curve. Therefore, the steps to solve this problem are:

1. Set the equation. Since both graphs have a common point, meaning they must equal.
2. Apply discriminant b²-4ac = 0 since tangent line and the curve only has one common point.

`\displaystyle{k - {x}^(2) = - 6x + 1}`

Arrange in quadratic terms:

`\displaystyle{ {x}^(2) - 6x + 1 - k = 0}`

Apply discriminant where b = -6, a = 1 and c = 1-k:

`\displaystyle{ {b}^(2) - 4ac = 0} \\ \\ \displaystyle{ {6}^(2) - 4(1)(1 - k) = 0} \\ \\ \displaystyle{ 36 - 4(1 - k)= 0} \\ \\ \displaystyle{ 36 - 4 + 4k = 0} \\ \\ \displaystyle{ 32 + 4k = 0} \\ \\ \displaystyle{4k = - 32} \\ \\ \displaystyle{k = - 8}`

Therefore the value of k that makes line tangent to the curve is -8.