Steve S. answered • 04/03/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
“on the same axes sketch the curves given by y=x^2( x - 4 ) and y = x ( 4 - x ). AND FIND THE coordinates of the point of intersection.
it is really confusing need your help badly…”
Break it down.
A. Graph two polynomial functions on the same coordinate system.
A.1. y = f(x) = x^2(x–4)
A.2. y = g(x) = x(4–x)
B. Find points of intersection (let’s do now):
x^2(x–4) = x(4–x)
x^3 – 4x^2 = 4x – x^2
x^3 – 3x^2 – 4x = 0
x(x^2 – 3x – 4) = 0
x(x – 4)(x + 1) = 0
x = –1,0,4
y = –1(4+1),0(4–0),4(4–4)
= –5,0,0
Points of intersection: (–1,–5),(0,0),(4,0)
Now let’s sketch the graphs:
A.1. y = f(x) = x^2(x–4)
This is a cubic polynomial function with end behavior up to the right (x → ∞ ==> f → ∞) and, because it has odd degree, down to the left.
Zeros: x = 0, multiplicity 2 (flattens and bounces off x axis); and x = 4.
<––––––0––––––0++++>f(x)
<––––––|––––––|––––––>x
0 4
There’s a relative maximum at x = 0 and a relative minimum between x = 0 and x = 4 where f’(x) = 0.
f(x) = x^2(x–4)
f’(x) = 2x(x–4) + x^2 = 3x^2 – 8x = x(3x–8)
Zeros of f’(x): x = 0, 8/3
Vertex of f’(x): x = (0 + 8/3)/2 = 4/3
(The vertex of f’ is where the change in f’ is 0, i.e., f’’(x) = 0. This point, (4/3,f(4/3)), is called the inflection point of f(x), and it marks the transition from f(x) opening up/down to opening down/up; the curvature of f(x).)
Relative maximum point of f(x): (0,0)
Relative minimum point of f(x): (8/3,f(8/3))
= (8/3,(8/3)^2(8/3–12/3))=(8/3,–256/27)≈(2.7,–9.5)
Inflection point of f(x): (4/3,f(4/3))
= (4/3,(16/9)(–8/3))=(4/3,–128/27)≈(1.3,–4.7)
Graph the points you know:
(–1,–5), (0,0), (1.3,–4.7), (2.7,–9.5), (4,0)
and draw a smooth curve through them with:
proper end behavior,
zero slope at extrema points, and
proper curvature.
A.2. y = g(x) = x(4–x)
This is a quadratic polynomial function with end behavior down to the right (x → ∞ ==> g → –∞) and, because it has even degree, down to the left. It’s a parabola opening downward.
Zeros: x = 0 and x = 4.
<––––––0+++++0––––––>f(x)
<––––––|–––––––|––––––>x
0 4
There’s a maximum when x = (4+0)/2 = 2.
Maximum point: (2,g(2)) = (2,4)
Axis of Symmetry: x = 2
(Use AoS to find corresponding points.)
Graph the points you know:
(–1,–5),(0,0),(2,4),(4,0),
and any new corresponding points:
(5,–5),
and draw a smooth curve through them.
When you have finished your graph, check it against http://www.wyzant.com/resources/files/267953/curve_sketching_intersecting_parabola_and_cubic.
it is really confusing need your help badly…”
Break it down.
A. Graph two polynomial functions on the same coordinate system.
A.1. y = f(x) = x^2(x–4)
A.2. y = g(x) = x(4–x)
B. Find points of intersection (let’s do now):
x^2(x–4) = x(4–x)
x^3 – 4x^2 = 4x – x^2
x^3 – 3x^2 – 4x = 0
x(x^2 – 3x – 4) = 0
x(x – 4)(x + 1) = 0
x = –1,0,4
y = –1(4+1),0(4–0),4(4–4)
= –5,0,0
Points of intersection: (–1,–5),(0,0),(4,0)
Now let’s sketch the graphs:
A.1. y = f(x) = x^2(x–4)
This is a cubic polynomial function with end behavior up to the right (x → ∞ ==> f → ∞) and, because it has odd degree, down to the left.
Zeros: x = 0, multiplicity 2 (flattens and bounces off x axis); and x = 4.
<––––––0––––––0++++>f(x)
<––––––|––––––|––––––>x
0 4
There’s a relative maximum at x = 0 and a relative minimum between x = 0 and x = 4 where f’(x) = 0.
f(x) = x^2(x–4)
f’(x) = 2x(x–4) + x^2 = 3x^2 – 8x = x(3x–8)
Zeros of f’(x): x = 0, 8/3
Vertex of f’(x): x = (0 + 8/3)/2 = 4/3
(The vertex of f’ is where the change in f’ is 0, i.e., f’’(x) = 0. This point, (4/3,f(4/3)), is called the inflection point of f(x), and it marks the transition from f(x) opening up/down to opening down/up; the curvature of f(x).)
Relative maximum point of f(x): (0,0)
Relative minimum point of f(x): (8/3,f(8/3))
= (8/3,(8/3)^2(8/3–12/3))=(8/3,–256/27)≈(2.7,–9.5)
Inflection point of f(x): (4/3,f(4/3))
= (4/3,(16/9)(–8/3))=(4/3,–128/27)≈(1.3,–4.7)
Graph the points you know:
(–1,–5), (0,0), (1.3,–4.7), (2.7,–9.5), (4,0)
and draw a smooth curve through them with:
proper end behavior,
zero slope at extrema points, and
proper curvature.
A.2. y = g(x) = x(4–x)
This is a quadratic polynomial function with end behavior down to the right (x → ∞ ==> g → –∞) and, because it has even degree, down to the left. It’s a parabola opening downward.
Zeros: x = 0 and x = 4.
<––––––0+++++0––––––>f(x)
<––––––|–––––––|––––––>x
0 4
There’s a maximum when x = (4+0)/2 = 2.
Maximum point: (2,g(2)) = (2,4)
Axis of Symmetry: x = 2
(Use AoS to find corresponding points.)
Graph the points you know:
(–1,–5),(0,0),(2,4),(4,0),
and any new corresponding points:
(5,–5),
and draw a smooth curve through them.
When you have finished your graph, check it against http://www.wyzant.com/resources/files/267953/curve_sketching_intersecting_parabola_and_cubic.
