Eric Nicholas K's answer to Nur Atika's Secondary 4 A Maths Singapore question.
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An idea. I assume the parabola is not a circle.
Date Posted:
4 years ago
Your 32/55 is missing an x
Good point
Alternative :
Both (0,0) and (16,0) are points of intersection with the line y = 0 and satisfy ax² + bx + c = 0
So x = 0 and x = 16 are the roots.
The midpoint of x = 0 and x = 16 is where the line of symmetry cuts the curve.
Midpoint is (0 + 16)/2 = 8
So x = 8 is the line of symmetry.
Vertex form of curve :
y = a(x - 8)² + k
Sub (0,0)
0 = a(-8)² + k
k = -64a
Sub (11,2)
2 = a(11 - 8)² + k
2 = 9a + k
Sub k = -64a,
2 = 9a - 64a
2 = -55a
a = -2/55
k = -64(-2/55) = 128/55
So y = -2/55 (x - 8)² + 128/55
y = -2/55 (x² - 16x + 64) + 128
y = -2/55 x² + 32/55 x
Both (0,0) and (16,0) are points of intersection with the line y = 0 and satisfy ax² + bx + c = 0
So x = 0 and x = 16 are the roots.
The midpoint of x = 0 and x = 16 is where the line of symmetry cuts the curve.
Midpoint is (0 + 16)/2 = 8
So x = 8 is the line of symmetry.
Vertex form of curve :
y = a(x - 8)² + k
Sub (0,0)
0 = a(-8)² + k
k = -64a
Sub (11,2)
2 = a(11 - 8)² + k
2 = 9a + k
Sub k = -64a,
2 = 9a - 64a
2 = -55a
a = -2/55
k = -64(-2/55) = 128/55
So y = -2/55 (x - 8)² + 128/55
y = -2/55 (x² - 16x + 64) + 128
y = -2/55 x² + 32/55 x
Very good and neat work wish he/she could help me more with my future question thank you!!
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