How do you use the limit definition to find the slope of the tangent line to the graph \[f\left( x \right)={{x}^{2}}-2x+4\] at \[\left( 0,4 \right)\] ?

Question

How do you use the limit definition to find the slope of the tangent line to the graph $$f\left( x \right)={{x}^{2}}-2x+4$$ at $$\left( 0,4 \right)$$ ?

Vedantu Content Team · Accepted Answer

Hint: We solve this question using a tangent line to a curve concept. We have to know the limit definition for the derivative formula to solve the given equation. We solve this problem using that formula. After that to get the slope evaluating the equation obtained at x=0 to get the solution. Complete step-by-step solution:Let us see the formula of limit definition of a derivative$$f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$$This is the derivative we use in this problem solving.Given equation$$f\left( x \right)={{x}^{2}}-2x+4$$Now we have to find the limit definition of the given equation.So we have the limit definition for a derivative formula as said above.The formula is$$f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$$Now we have to substitute the given question in the formula above. We have made $$x$$ as $$x+h$$ while substituting in place of $$f\left( x+h \right)$$. By substituting the equation in the formula we will get$$f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{{{\left( x+h \right)}^{2}}-2\left( x+h \right)+4-{{x}^{2}}+2x-4}{h}$$Now we have to simplify it.First we have to expand the first term.$$\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{x}^{2}}+2hx+{{h}^{2}}-2\left( x+h \right)-{{x}^{2}}+2x}{h}$$Now we have to multiply the second term with 2.$$\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{{{x}^{2}}+2hx+{{h}^{2}}-2x-2h-{{x}^{2}}+2x}{h}$$Now we have to simplify it by subtracting like terms. We will get$$\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{2hx}{h}-\dfrac{2h}{h}+\dfrac{{{h}^{2}}}{h}$$Now we have to simplify the terms $$\Rightarrow \displaystyle \lim_{h \to 0}2x-2+h$$Now we have to take out the terms which didn’t contain h in them. We will get$$\Rightarrow 2x-2+\displaystyle \lim_{h \to 0}h$$Now by applying limits we will get$$\Rightarrow 2x-2$$So the derivative of limit definition we got is $$f'\left( x \right)=2x-2$$Now We have  found the gradient of the tangent at any point. To get the slope we have to evaluate it at x=0. By substituting we will get$$f'\left( 0 \right)=2\left( 0 \right)-2$$We will get$$\Rightarrow -2$$ So the slope of the tangent of the given line is $$-2$$.Note:  We have to be aware of the formula while solving this question. We have to be careful while applying limits and taking terms out this may result in change of the solution. Also we had to know to find the tangent we have found at any point and to get the slope of a particular point we have substituted its x coordinate.