Sketch the graph of the function y = x3 - 2x2 - 3x. Find the intercepts and turning points.
|Mathematics
Sketch the graph of the function y = x3 - 2x2 - 3x. Find the intercepts and turning points.
ScanSolve AI Answer
5 steps
Answer
(-1, 0), (0, 0), and (3, 0)
You're on a roll —
To sketch the graph of y=x3−2x2−3x, we need to find the intercepts, turning points, and the point of inflection.
Continuing from 9.1: Determine the intercepts with the axes.
Step 2 (continued): Factor the quadratic x2−2x−3=0.
We need two numbers that multiply to -3 and add to -2. These are -3 and 1.
x(x−3)(x+1)=0
This gives us the x-intercepts:
x=0orx−3=0orx+1=0x=0orx=3orx=−1
The x-intercepts are ∗(−1,0),(0,0),and(3,0)∗
9.2 Determine the coordinates of the turning points.
Step 3: Find the first derivative of the function.
y=x3−2x2−3xdxdy=3x2−4x−3
Step 4: Set the first derivative to zero to find the x-coordinates of the turning points.
3x2−4x−3=0
Use the quadratic formula x=2a−b±b2−4ac:
x=2(3)−(−4)±(−4)2−4(3)(−3)x=64±16+36x=64±52x=64±213x=32±13
This gives two x-coordinates for the turning points:
x1=32+13≈32+3.605≈35.605≈1.868x2=32−13≈32−3.605≈3−1.605≈−0.535
Step 5: Substitute these x-values back into the original function to find the corresponding y-coordinates.
For x1≈1.868:
y1=(1.868)3−2(1.868)2−3(1.868)y1≈6.505−2(3.490)−5.604y1≈6.505−6.980−5.604y1≈−6.079
Turning Point 1: ∗(1.87,−6.08)∗ (local minimum)
For x2≈−0.535:
y2=(−0.535)3−2(−0.535)2−3(−0.535)y2≈−0.153−2(0.286)+1.605y2≈−0.153−0.572+1.605y2≈0.880
Turning Point 2: ∗(−0.54,0.88)∗ (local maximum)
9.3 Determine the coordinates of the point of inflection.
Step 6: Find the second derivative of the function.
dxdy=3x2−4x−3dx2d2y=6x−4
Step 7: Set the second derivative to zero to find the x-coordinate of the point of inflection.
6x−4=06x=4x=64=32
Step 8: Substitute this x-value back into the original function to find the corresponding y-coordinate.
y=(32)3−2(32)2−3(32)y=278−2(94)−2y=278−98−2
To combine these, find a common denominator (27):
y=278−9×38×3−1×272×27y=278−2724−2754y=278−24−54y=27−70y≈−2.59
The point of inflection is \boxed{*(\frac{2{3}, -\frac{70}{27}) \approx (0.67, -2.59)*}}
9.4 Sketch the graph.
To sketch the graph, plot the following points:
x-intercepts:(−1,0), (0,0), (3,0)
y-intercept:(0,0)
Local maximum:(−0.54,0.88)
Local minimum:(1.87,−6.08)
Point of inflection:(0.67,−2.59)
Since it's a cubic function with a positive leading coefficient (x3), the graph will:
Start from negative infinity on the left (as x→−∞,y→−∞).
Rise to the local maximum at (−0.54,0.88).
Decrease, passing through the y-intercept (0,0) and the point of inflection (0.67,−2.59), to the local minimum at (1.87,−6.08).
Increase from the local minimum, passing through the x-intercept (3,0), towards positive infinity on the right (as x→∞,y→∞).
A sketch would look like this:
(Imagine an x-y coordinate plane)
Mark the x-intercepts at -1, 0, and 3.
Mark the y-intercept at 0.
Plot the local maximum slightly to the left of the y-axis and above it.
Plot the local minimum to the right of the y-axis and significantly below the x-axis.
Draw a smooth curve connecting these points, starting from the bottom left, going up through (−1,0) to the local max, then down through (0,0) and the point of inflection to the local min, and finally up through (3,0) to the top right.
[Since I cannot draw an image, I have provided a detailed description of how to sketch the graph based on the calculated points.]
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ScanSolve AI Answer
You're on a roll — To sketch the graph of y = x^3 - 2x^2 - 3x, we need to find the intercepts, turning points, and the point of inflection. Continuing from 9.1: Determine the intercepts with the axes. Step 2 (continued): Factor the quadratic x^2 - 2x - 3 = 0. We need two numbers that multiply to -3 and add to -2. These are -3 and 1. x(x-3)(x+1) = 0 This gives us the x-intercepts: x = 0 or x - 3 = 0 or x + 1 = 0 x = 0 or x = 3 or x = -1 The x-intercepts are (-1, 0), (0, 0), and (3, 0) 9.2 Determine the coordinates of the turning points. Step 3: Find the first derivative of the function. y = x^3 - 2x^2 - 3x (dy)/(dx) = 3x^2 - 4x - 3 Step 4: Set the first derivative to zero to find the x-coordinates of the turning points. 3x^2 - 4x - 3 = 0 Use the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a: x = -(-4) ± sqrt((-4)^2 - 4(3)(-3))2(3) x = 4 ± sqrt(16 + 36)6 x = 4 ± sqrt(52)6 x = 4 ± 2sqrt(13)6 x = 2 ± sqrt(13)3 This gives two x-coordinates for the turning points: x_1 = 2 + sqrt(13)3 ≈ (2 + 3.605)/(3) ≈ (5.605)/(3) ≈ 1.868 x_2 = 2 - sqrt(13)3 ≈ (2 - 3.605)/(3) ≈ (-1.605)/(3) ≈ -0.535 Step 5: Substitute these x-values back into the original function to find the corresponding y-coordinates. For x_1 ≈ 1.868: y_1 = (1.868)^3 - 2(1.868)^2 - 3(1.868) y_1 ≈ 6.505 - 2(3.490) - 5.604 y_1 ≈ 6.505 - 6.980 - 5.604 y_1 ≈ -6.079 Turning Point 1: (1.87, -6.08) (local minimum) For x_2 ≈ -0.535: y_2 = (-0.535)^3 - 2(-0.535)^2 - 3(-0.535) y_2 ≈ -0.153 - 2(0.286) + 1.605 y_2 ≈ -0.153 - 0.572 + 1.605 y_2 ≈ 0.880 Turning Point 2: (-0.54, 0.88) (local maximum) 9.3 Determine the coordinates of the point of inflection. Step 6: Find the second derivative of the function. (dy)/(dx) = 3x^2 - 4x - 3 (d^2y)/(dx^2) = 6x - 4 Step 7: Set the second derivative to zero to find the x-coordinate of the point of inflection. 6x - 4 = 0 6x = 4 x = (4)/(6) = (2)/(3) Step 8: Substitute this x-value back into the original function to find the corresponding y-coordinate. y = ((2)/(3))^3 - 2((2)/(3))^2 - 3((2)/(3)) y = (8)/(27) - 2((4)/(9)) - 2 y = (8)/(27) - (8)/(9) - 2 To combine these, find a common denominator (27): y = (8)/(27) - (8 × 3)/(9 × 3) - (2 × 27)/(1 × 27) y = (8)/(27) - (24)/(27) - (54)/(27) y = (8 - 24 - 54)/(27) y = (-70)/(27) y ≈ -2.59 The point of inflection is ((2)/(3), -(70)/(27)) ≈ (0.67, -2.59) 9.4 Sketch the graph. To sketch the graph, plot the following points: x-intercepts: (-1, 0), (0, 0), (3, 0) y-intercept: (0, 0) Local maximum: (-0.54, 0.88) Local minimum: (1.87, -6.08) Point of inflection: (0.67, -2.59) Since it's a cubic function with a positive leading coefficient (x^3), the graph will: Start from negative infinity on the left (as x -, y -). Rise to the local maximum at (-0.54, 0.88). Decrease, passing through the y-intercept (0,0) and the point of inflection (0.67, -2.59), to the local minimum at (1.87, -6.08). Increase from the local minimum, passing through the x-intercept (3,0), towards positive infinity on the right (as x , y ). A sketch would look like this: (Imagine an x-y coordinate plane) 1. Mark the x-intercepts at -1, 0, and 3. 2. Mark the y-intercept at 0. 3. Plot the local maximum slightly to the left of the y-axis and above it. 4. Plot the local minimum to the right of the y-axis and significantly below the x-axis. 5. Draw a smooth curve connecting these points, starting from the bottom left, going up through (-1,0) to the local max, then down through (0,0) and the point of inflection to the local min, and finally up through (3,0) to the top right. [Since I cannot draw an image, I have provided a detailed description of how to sketch the graph based on the calculated points.] What's next?