Find the equation of the tangent to the curve $y=x^2$ at the point $(2,4)$.

For $y=x^2$, the derivative (power rule, Section 10) is $y'=2x$, so the slope of the tangent at $x=2$ is $y'(2)=2(2)=4$ (Section 7). The point on the curve is $(2,4)$. Using the point-slope form $y-y_1=m(x-x_1)$: $$y-4 = 4(x-2) \quad\Longrightarrow\quad y = 4x-8+4 = 4x-4.$$ > [!ANSWER] $y = 4x-4$

Find the equation of the tangent to the curve y=x²

Concept understanding — Derivative as Slope of the Tangent

For a curve $y=f(x)$ and a fixed point

$A=(a,f(a))$ on it, a nearby second point $B=(a+h,f(a+h))$ determines a secant line through $A$

and $B$ whose slope is the difference quotient $[f(a+h)-f(a)]/h$ -- the same expression that

defines the derivative algebraically. As $h\to0$ , point $B$ slides along the curve toward $A$ ,

and the secant line rotates toward a limiting position called the tangent to the curve at $A$ ;

the slope of this tangent is therefore exactly $\lim_{h\to0}[f(a+h)-f(a)]/h=f'(a)$ . This shows

that the algebraic definition of the derivative (instantaneous rate of change) and its geometric

meaning (slope of the tangent) are not two separate facts that happen to coincide, but literally

the same limit viewed two different ways -- and once $f'(a)$ is known, the equation of the

tangent line follows immediately from the point-slope form, $y-f(a)=f'(a)(x-a)$ .

Loading…