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# 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 cofusing need ur help badly...

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Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...
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“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.

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.