Answer:
[tex]\textsf{1)} \quad y=x+\dfrac{17}{4}[/tex]
[tex]\textsf{2)} \quad y=2x+4[/tex]
Step-by-step explanation:
Differentiation is an algebraic process that finds the gradient (slope) of a curve.
At a point, the gradient of a curve is the same as the gradient of the tangent line to the curve at that point.
Given function:
[tex]y=4+2x-x^2[/tex]
Differentiate the given function:
[tex]\implies \dfrac{\text{d}y}{\text{d}x}=2-2x[/tex]
If the tangent to the parabola forms an angle of 45° with the negative direction of the x-axis, then the gradient of the tangent is 1.
Set the differentiated function to 1 to find the x-value of the point where the gradient of the tangent is 1:
[tex]\implies \dfrac{\text{d}y}{\text{d}x}=1[/tex]
[tex]\implies 2-2x=1[/tex]
[tex]\implies -2x=-1[/tex]
[tex]\implies x=\dfrac{1}{2}[/tex]
Substitute x = 1/2 into the given function to find the y-value of the point:
[tex]\implies y=4+2\left(\dfrac{1}{2}\right)-\left(\dfrac{1}{2}\right)^2=\dfrac{19}{4}[/tex]
Therefore, the point at which the tangent of the parabola forms an angle of 45° with the negative direction of the x-axis is:
- [tex]\left(\dfrac{1}{2},\dfrac{19}{4}\right)[/tex]
Substitute the found point and slope into the point-slope formula to create an equation of the tangent:
[tex]\implies y-y_1=m(x-x_1)[/tex]
[tex]\implies y-\dfrac{19}{4}=1\left(x-\dfrac{1}{2}\right)[/tex]
[tex]\implies y=x+\dfrac{17}{4}[/tex]
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Given function:
[tex]y=4+2x-x^2[/tex]
Differentiate the given function:
[tex]\implies \dfrac{\text{d}y}{\text{d}x}=2-2x[/tex]
If the tangent to the parabola is parallel to the line y = 2x - 4 then the gradient of the tangent is 2 (since parallel lines have the same gradient).
Set the differentiated function to 2 to find the x-value of the point where the gradient of the tangent is 2:
[tex]\implies \dfrac{\text{d}y}{\text{d}x}=2[/tex]
[tex]\implies 2-2x=2[/tex]
[tex]\implies -2x=0[/tex]
[tex]\implies x=0[/tex]
Substitute x = 0 into the given function to find the y-value of the point:
[tex]\implies y=4+2\left(0\right)-\left(0\right)^2=4[/tex]
Therefore, the point at which the tangent of the parabola is parallel to the line y = 2x - 4 is:
Substitute the found point and slope into the point-slope formula to create an equation of the tangent:
[tex]\implies y-y_1=m(x-x_1)[/tex]
[tex]\implies y-4=2\left(x-0\right)[/tex]
[tex]\implies y=2x+4[/tex]
