Question:
What is the equation of the tangent line to the function y=2x^2 - x + 3 at x= 1
a. y=4x+8
b. y=4x-8
c. y=4x-6
d. y=4x
Answer:
The equation of tangent is y = 3 x+1
Step-by-step explanation:
Given the equation of the tangent line to the function [tex]y=2 x^{2}-x+3[/tex]
Need to find out the equation of tangent at x = 1
We know that the equation of tangent for any function at point A is
[tex]\frac{y-y_{1}}{x-x_{1}}=\text {slope of line at point } A[/tex]
According to question, [tex]x_{1}=1[/tex]
To find out the value of [tex]y_{1}[/tex], we put the value of [tex]x_{1}[/tex] in the given equation as,
[tex]y_{1}=2\left(x_{1}\right)^{2}-\left(x_{1}\right)+3=2 \times(1)^{2}-1+3=4[/tex]
Slope of line at x=1 is
[tex]\text {slope}=4 x_{1}-1=4 \times 1-1=3[/tex]
So the required equation of tangent is
[tex]\frac{y-4}{x-1}=3[/tex]
[tex]y-4=3 x-3[/tex]
[tex]y=3 x+1[/tex]
