Finding a downward sloping function given a tangent point on a linear curve

Question

I have a linear function $10-5x$. I am trying to create a downward sloping curve at a point where the tangency point is $(1.5, 2.5)$.

I tried using a function of $3.75/x$, but instead of making it a tangent to the linear curve it instead intersects it. How do I make a function that is tangent to the linear curve at point $(1.5,2.5)$?

Answer 1

HINT

Let try with a parabola $y=ax^2+bx+c$ with the conditions

• passing through the point $(1.5,2.5)$: $2.5=a(1.5)^2+b(1.5)+c$
• same tangent at the point $(1.5,2.5)$: $-5=2a(1.5)+b$
• open upward: $a>0$ (free parameter)

Answer 2

Lets continue with what you had in mind. You wanted to fit a hyperbola to be the tangent to the line. You need at the point x=1.5 to have gradient -5 and you also need for (1.5,2.5) to be a solution. We begin with 5*(1.5)^2/x to get the gradient that you want of -5 at the point x=1.5. We can then adjust this function on the y-axis by adding or subtracting some amount to make it a tangent. The final curve of y = 11.25/x-5 fits what you wanted.

Answer 3

Saying that the tangent line at $P(1.5,2.5)$ has equation $y=10-5x$ gives you two conditions about a function $f$:

1. $P$ is on the graph of $f$, that is $f(1.5)=2.5$
2. The tangent at $P$ has slope $-5$, that is $f'(1.5)=-5$.

You tried a function of the form $f(x)=\frac{a}{x}$. There is only one parameter so it was unlikely to lead to a solution.

Try functions with two or more parameters: $f(x)=\frac{a}{x}+b$, $f(x)=\frac{a}{x+b}$, etc.