find the equation of the line tangent to the function at the given point. y=x^3-2x^2+4 at (1,3)

[tex]\begin{gathered} y=x^3-2x^2\text{ + 4} \\ \frac{d\text{ y}}{d\text{ x}}=y^{^{\prime}}=3x^{3-1}\text{ - 2}\cdot2x^{2-1}\text{ + }0 \\ y^{^{\prime}}=3x^2\text{ - 4x }\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots..\text{ (2)} \end{gathered}[/tex]

Step 2: To determine the slope (m) of the tangent line, insert the x-value (1) into the equation 2

[tex]\begin{gathered} y^{^{\prime}}=3x^2\text{ - 4x} \\ y^{^{\prime}}=3(1)^2^{}\text{ - 4}(1) \\ y^{^{\prime}}=\text{ 3 - 4} \\ y^{^{\prime}}=\text{ - 1} \end{gathered}[/tex]

Step 3: use the point-slope formula to determine the equation of the line tangent to the given function at x = 1

[tex]\begin{gathered} y\text{ - }y_1=m(x-x_1) \\ y\text{ - 3 = -1(x - 1)} \\ y\text{ - 3 = -x + 1} \\ y\text{ = -x + 1 + 3} \\ y\text{ = -x + 4 } \\ \end{gathered}[/tex]

Hence, The equation of the line tangent to the function at the given point is:

[tex]y\text{ = -x + 4 }[/tex]

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[tex]y=(-3x+6)^{\frac{1}{2}}[/tex]

1. Take 1st derivative

[tex]\frac{d\text{ y}}{d\text{ x }}=y^{^{\prime}}\text{ = }\frac{1}{2}(-3)(-3x+6)^{-\frac{1}{2}}[/tex][tex]y^{^{\prime}}\text{ = -}\frac{3}{2}(-3x+6)^{-\frac{1}{2}}[/tex]

2. insert the x-value (-1) to determine the slope (m)

[tex]\begin{gathered} y^{^{\prime}}\text{ = -}\frac{3}{2}(-3(-1)+6)^{-\frac{1}{2}} \\ y^{^{\prime}}\text{ = -}\frac{3}{2}(9)^{-\frac{1}{2}} \\ y^{^{\prime}}\text{ = -}\frac{3}{2}(\frac{1}{\sqrt[]{9}}) \\ y^{^{\prime}}\text{ = -}\frac{3}{2}\text{ }\cdot\text{ }\frac{1}{3} \\ y^{^{\prime}}\text{ = -}\frac{1}{2} \end{gathered}[/tex]

3. Now, determine the equation of the line tangent at (-1,3)

[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y\text{ - 3 = -}\frac{1}{2}(x\text{ + 1)} \\ y\text{ = -}\frac{x}{2}\text{ - }\frac{1}{2}\text{ + 3} \\ y\text{ = -}\frac{x}{2}\text{ + }\frac{5}{2} \end{gathered}[/tex]