Answer :
To solve the problem, we will proceed step by step:
### (a) The Slope of the Curve
First, we need to find the slope of the curve at the given point [tex]\( x = -1 \)[/tex] for the curve [tex]\( y = -x^2 \)[/tex].
1. Find the derivative of [tex]\( y = -x^2 \)[/tex]:
The derivative [tex]\( y' \)[/tex] of [tex]\( y = -x^2 \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ y' = \frac{d(-x^2)}{dx} = -2x \][/tex]
2. Evaluate the derivative at [tex]\( x = -1 \)[/tex]:
[tex]\[ y'(-1) = -2(-1) = 2 \][/tex]
So, the slope of the curve at [tex]\( x = -1 \)[/tex] is [tex]\( 2 \)[/tex].
### (b) Equation of the Tangent Line
To find the equation of the tangent line at [tex]\( x = -1 \)[/tex], we need the slope we calculated and the point where the tangent touches the curve.
1. Find the [tex]\( y \)[/tex]-coordinate of the point at [tex]\( x = -1 \)[/tex]:
Substitute [tex]\( x = -1 \)[/tex] into the original equation [tex]\( y = -x^2 \)[/tex]:
[tex]\[ y = -(-1)^2 = -1 \][/tex]
So, the point of tangency is [tex]\( (-1, -1) \)[/tex].
2. Use the point-slope form of the line equation:
The point-slope form of the equation of a line is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope, and [tex]\((x_1, y_1)\)[/tex] is the point of tangency.
[tex]\[ y - (-1) = 2(x - (-1)) \][/tex]
Simplify:
[tex]\[ y + 1 = 2(x + 1) \][/tex]
[tex]\[ y + 1 = 2x + 2 \][/tex]
[tex]\[ y = 2x + 1 \][/tex]
So, the equation of the tangent line is [tex]\( y = 2x + 1 \)[/tex].
### (c) Equation of the Normal Line
The normal line is perpendicular to the tangent line at the point of tangency. Therefore, its slope is the negative reciprocal of the tangent slope.
1. Slope of the normal line:
The slope of the tangent line is [tex]\( 2 \)[/tex], so the slope of the normal line is:
[tex]\[ m = -\frac{1}{2} \][/tex]
2. Use the point-slope form of the line equation:
Again, use the point-slope form with the point of tangency [tex]\((-1, -1)\)[/tex]:
[tex]\[ y - (-1) = -\frac{1}{2}(x - (-1)) \][/tex]
Simplify:
[tex]\[ y + 1 = -\frac{1}{2}(x + 1) \][/tex]
[tex]\[ y + 1 = -\frac{1}{2}x - \frac{1}{2} \][/tex]
[tex]\[ y = -\frac{1}{2}x - \frac{1}{2} - 1 \][/tex]
[tex]\[ y = -\frac{1}{2}x - \frac{3}{2} \][/tex]
So, the equation of the normal line is [tex]\( y = -\frac{1}{2}x - \frac{3}{2} \)[/tex].
### (d) Graph of the Curve, Tangent Line, and Normal Line
To visualize these equations on a graph, we will plot:
1. The original curve [tex]\( y = -x^2 \)[/tex].
2. The tangent line [tex]\( y = 2x + 1 \)[/tex].
3. The normal line [tex]\( y = -\frac{1}{2}x - \frac{3}{2} \)[/tex].
#### Step-by-step to make the graph:
1. Choose a range for [tex]\( x \)[/tex] that includes the point [tex]\( x = -1 \)[/tex].
2. Plot the curve [tex]\( y = -x^2 \)[/tex].
3. Plot the tangent line [tex]\( y = 2x + 1 \)[/tex].
4. Plot the normal line [tex]\( y = -\frac{1}{2}x - \frac{3}{2} \)[/tex].
#### Graph example:
- Let [tex]\( x \)[/tex] range from [tex]\(-2\)[/tex] to [tex]\( 2 \)[/tex].
Plot these lines along with the curve, and ensure you mark the point [tex]\((-1, -1)\)[/tex] clearly to show where the tangency and normal intersect the curve.
By following these steps, you will be able to graphically represent the equations and observe how the tangent and normal lines interact with the curve at the given point.
### (a) The Slope of the Curve
First, we need to find the slope of the curve at the given point [tex]\( x = -1 \)[/tex] for the curve [tex]\( y = -x^2 \)[/tex].
1. Find the derivative of [tex]\( y = -x^2 \)[/tex]:
The derivative [tex]\( y' \)[/tex] of [tex]\( y = -x^2 \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ y' = \frac{d(-x^2)}{dx} = -2x \][/tex]
2. Evaluate the derivative at [tex]\( x = -1 \)[/tex]:
[tex]\[ y'(-1) = -2(-1) = 2 \][/tex]
So, the slope of the curve at [tex]\( x = -1 \)[/tex] is [tex]\( 2 \)[/tex].
### (b) Equation of the Tangent Line
To find the equation of the tangent line at [tex]\( x = -1 \)[/tex], we need the slope we calculated and the point where the tangent touches the curve.
1. Find the [tex]\( y \)[/tex]-coordinate of the point at [tex]\( x = -1 \)[/tex]:
Substitute [tex]\( x = -1 \)[/tex] into the original equation [tex]\( y = -x^2 \)[/tex]:
[tex]\[ y = -(-1)^2 = -1 \][/tex]
So, the point of tangency is [tex]\( (-1, -1) \)[/tex].
2. Use the point-slope form of the line equation:
The point-slope form of the equation of a line is [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( m \)[/tex] is the slope, and [tex]\((x_1, y_1)\)[/tex] is the point of tangency.
[tex]\[ y - (-1) = 2(x - (-1)) \][/tex]
Simplify:
[tex]\[ y + 1 = 2(x + 1) \][/tex]
[tex]\[ y + 1 = 2x + 2 \][/tex]
[tex]\[ y = 2x + 1 \][/tex]
So, the equation of the tangent line is [tex]\( y = 2x + 1 \)[/tex].
### (c) Equation of the Normal Line
The normal line is perpendicular to the tangent line at the point of tangency. Therefore, its slope is the negative reciprocal of the tangent slope.
1. Slope of the normal line:
The slope of the tangent line is [tex]\( 2 \)[/tex], so the slope of the normal line is:
[tex]\[ m = -\frac{1}{2} \][/tex]
2. Use the point-slope form of the line equation:
Again, use the point-slope form with the point of tangency [tex]\((-1, -1)\)[/tex]:
[tex]\[ y - (-1) = -\frac{1}{2}(x - (-1)) \][/tex]
Simplify:
[tex]\[ y + 1 = -\frac{1}{2}(x + 1) \][/tex]
[tex]\[ y + 1 = -\frac{1}{2}x - \frac{1}{2} \][/tex]
[tex]\[ y = -\frac{1}{2}x - \frac{1}{2} - 1 \][/tex]
[tex]\[ y = -\frac{1}{2}x - \frac{3}{2} \][/tex]
So, the equation of the normal line is [tex]\( y = -\frac{1}{2}x - \frac{3}{2} \)[/tex].
### (d) Graph of the Curve, Tangent Line, and Normal Line
To visualize these equations on a graph, we will plot:
1. The original curve [tex]\( y = -x^2 \)[/tex].
2. The tangent line [tex]\( y = 2x + 1 \)[/tex].
3. The normal line [tex]\( y = -\frac{1}{2}x - \frac{3}{2} \)[/tex].
#### Step-by-step to make the graph:
1. Choose a range for [tex]\( x \)[/tex] that includes the point [tex]\( x = -1 \)[/tex].
2. Plot the curve [tex]\( y = -x^2 \)[/tex].
3. Plot the tangent line [tex]\( y = 2x + 1 \)[/tex].
4. Plot the normal line [tex]\( y = -\frac{1}{2}x - \frac{3}{2} \)[/tex].
#### Graph example:
- Let [tex]\( x \)[/tex] range from [tex]\(-2\)[/tex] to [tex]\( 2 \)[/tex].
Plot these lines along with the curve, and ensure you mark the point [tex]\((-1, -1)\)[/tex] clearly to show where the tangency and normal intersect the curve.
By following these steps, you will be able to graphically represent the equations and observe how the tangent and normal lines interact with the curve at the given point.
