# What is the tangent line of f(x) = $x^2 + 1$ and perpendicular to the line $y = -x + 4$? [closed]

I'm having difficult to determine the tangent line of $$f(x) = x^2 + 1$$ which is perpendicular to the line $$y = -x + 4$$. Could someone help me?

## closed as off-topic by José Carlos Santos, John Douma, mrtaurho, mihaild, ShaileshJul 6 at 0:01

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• Let $y=mx+h$ is desired line, then $m=1$. this line intersect the curve in one point. – Nosrati Jul 5 at 16:10
• Ermm... what is $t$ here? – amd Jul 5 at 23:27

The tangent line of a curve $$f$$ in $$(a,f(a))$$ is the line $$g(x) = f'(a)(x-a) + f(a)$$

So we have that $$g$$ has to be perpendicular to $$y = -t + 4$$

Then $$f'(a) = 1$$, thus we have $$f'(a) = 2a = 1 \implies a = \frac{1}{2}$$

Then the tangent line is $$g(x) = f'(\frac{1}{2}) (x-\frac{1}{2}) + f(\frac{1}{2}) = (x-\frac{1}{2}) + \frac{5}{4}$$

First you find the gradient of the required line. Perpendicular lines have slopes whose product is $$-1$$, so the gradient of the line is $$1$$ and its equation can be written as $$y=x+c$$.

You can now proceed in one of two ways.

The first is calculus - find the derivative of the function of the curve, etc. which has already been mentioned in another answer.

An alternative pre-calculus way is to use the theory of quadratics.

The line is tangent to the curve so there is exactly one point in common. Hence there is only one solution for $$x^2+1 = x+c$$ which is $$x^2 - x + (1-c)=0$$.

Set the discriminant to zero and solve: $$1-4(1-c)=0$$ giving $$c=\frac 34$$. So the required equation is $$y=x + \frac34$$.