Respuesta :
Equation of a Line
The equation of a line that passes through the point (h, k) and has a slope m, is given by:
[tex]y-k=m(x-h)[/tex]This is known as the point-slope form of the line.
We already know the coordinates of the point (1, 1) but we don't know the value of the slope m. We will find it out by using the rest of the data.
Our line is perpendicular to the function:
[tex]f(x)=x^2+2x-2[/tex]At the given point. To find the slope of the tangent line, we use derivatives:
[tex]f^{\prime}(x)=2x+2[/tex]Substitute x = 1:
[tex]\begin{gathered} f^{\prime}(1)=2\cdot1+2 \\ f^{\prime}(1)=4 \end{gathered}[/tex]Now we know the slope of the tangent line, but our line is perpendicular to that line, so we find the perpendicular slope with the formula:
[tex]\begin{gathered} m_2=-\frac{1}{m} \\ m_2=-\frac{1}{4} \end{gathered}[/tex]We're ready to find the required equation. Substituting the coordinates of the point and the just-found slope:
[tex]y-1=-\frac{1}{4}(x-1)[/tex]This is the point-slope form, but maybe it's required to find the slope-intercept form. Multiply by 4:
[tex]\begin{gathered} 4y-4=-x+1 \\ \text{Add 4:} \\ 4y=-x+5 \\ \text{Divide by 4:} \\ y=-\frac{1}{4}x+\frac{5}{4} \end{gathered}[/tex]This is the answer is slope-intercept form
