Consider the following expression:
[tex]y^2=x^3+3x^2[/tex]In the first place, we differentiate this expression using implicit differentiation. For this, we treat y as a function of x. Therefore, we get:
[tex]2y\text{ }\frac{dy}{dx}=3x^2+6x[/tex]solving for dy/dx, we get:
Equation 1:
[tex]y^{\prime}(x)=\frac{dy}{dx}=\frac{3x^2+6x}{2y}[/tex]Now, remember that the equation of the tangent line is given by the following formula:
Equation 2:
[tex]y=y_0+y^{\prime}(x_0)(x\text{ - x}_0)[/tex]If (x_0 , y_0 ) = (1 , 2), the above equation becomes:
[tex]y=2+y^{\prime}(x_0)(x\text{ - 1})[/tex]Now, applying the equation 1 using the point (x_0 , y_0 ) = (1 , 2), we get the slope of the line at this point:
[tex]y^{\prime}(x_0)=\frac{3(x_0)^2+6(x_0)}{2y_0}=\frac{3(1)^2+6(1)}{2(2)}=\frac{3+6}{4}=\frac{9}{4}[/tex]applying this value in equation 2, we get:
[tex]y=2+\frac{9}{4}(x\text{ - 1})[/tex]this is equivalent to:
Equation 3:
[tex]y\text{ - 2}=\frac{9}{4}(x\text{ - 1})[/tex]Now, remember that the Point-Slope Form of a Line is given by the following formula:
[tex]y\text{ - y}_0=m(x\text{ - x}_0)[/tex]we can see that equation 3 is expressed in that way. Thus, we can conclude that the correct answer is:
Answer:[tex]y\text{ - 2}=\frac{9}{4}(x\text{ - 1})[/tex]