Basic Parabola Practice

Basic Parabola Practice

1) General to Standard Form

Write these rules in standard form, then describe the curve by listing:

 i) vertex coordinates ii) equation of the axis of symmetry iii) max or min value iv) domain v) range a) f (x) = x 2 + 6x + 8 b) g (x) = –3x 2 + 12x – 10 c) h (x) = 2x 2 + 5x – 3 d) k (x) = x 2 – 4

(solution)

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2) Find Rule of Correspondence

Find the rule of correspondence for a parabola:
a) with vertex at (1, 1) and y-intercept of 5.
b) with zeros at –1 and 3, passing through P(7, –16).

(solution)

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3) Find Zeros

Find the zeros for these parabolas.
a) f (x) = 2x 2 + 5x – 3
b) f (x) = – (x + 1) 2 + 4
c) f (x) = –5x 2 + 4x + 1
d) f (x) = 2(x 2 – 8x + 17)

(solution)

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Solutions

Write the following rules in standard form, then describe the curve by listing:

 a) f (x) = x 2 + 6x + 8 becomes f (x) = (x + 3)² – 1 i) vertex: (–3, –1) iii) minimum = –1 ii) axis: x = –3 iv) domain: R; v) range: b) g (x) = –3x 2 + 12x – 10 g (x) = –3(x – 2)² + 2 i) vertex: (2, 2) iii) maximum = 2 ii) axis: x = 2 iv) domain: R; v) range: . c) h (x) = 2x 2 + 5x – 3 h (x) = 2(x + 5/4)² – 49/8 i) vertex: (–5/4, – 49/8) iii) minimum = – 49/8 ii) axis: x = –5/4 iv) domain: R; v) range: . d) k (x) = x 2 – 4 i) vertex: (0, – 4) iii) minimum = – 4 ii) axis: x = 0, the y-axis iv) domain: R; v) range: .

a) with vertex at (1, 1) and y-intercept of 5.

We know that h = 1, k = 1, and P (0, 5). We use standard form:
So f (x) = a (x – h)² + k becomes f (x) = a (x – 1)² + 1
Now we substitute x = 0 and f (x) = 5 from point P, so: 5 = a (0 – 1)² + 1

This says that a = 4. Therefore, the rule is: f (x) = 4 (x – 1)² + 1

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b) with zeros at –1 and 3, passing through P(7, –16).

Let's use the zeros form for this one: g (x) = a(x – x1)(x – x2)
So, g (x) = a(x + 1)(x – 3), and substituting for P we get: –16 = a(7 – 3)(7 + 1)

This says that a = – ½. Therefore, the rule is: g (x) = –½ (x – 3)(x + 1)

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 a) f (x) = 2x 2 + 5x – 3 0 = (2x – 1)(x + 3) x = ½ or x = –3 b) f (x) = – (x + 1) 2 + 4 (x + 1) 2 = 4 so x + 1 = ± 2 x = 1 or x = –3 c) f (x) = –5x 2 + 4x + 1 0 = (1 + 5x)(1 – x) x = 1 or x = – 1/5 d) f (x) = 2(x 2 – 8x + 17) b² – 4ac = – 4 no Real solutions

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