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) = x² + 6x + 8 b) g (x) = –3x² + 12x – 10

c) h (x) = 2x² + 5x – 3 d) k (x) = x² – 4

(solution)

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)

3) Find Zeros

Find the zeros for these parabolas.

(solution)

Solutions

1) General to Standard Form

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

a) f (x) = x² + 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) = –3x² + 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) = 2x² + 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) = x² – 4 i) vertex: (0, – 4)
iii) minimum = – 4 ii) axis: x = 0, the y-axis
iv) domain: R; v) range: .

2) Find Rule of Correspondence

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

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)

3) Find the zeros for these parabolas.

a) f (x) = 2x² + 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) = –5x² + 4x + 1 0 = (1 + 5x)(1 – x) x = 1 or x = – 1/5

d) f (x) = 2(x² – 8x + 17) b² – 4ac = – 4 no Real solutions

( MathTub Index )

i) vertex coordinates	ii) equation of the axis of symmetry	iii) max or min value
iv) domain	v) range
a) f (x) = x² + 6x + 8	b) g (x) = –3x² + 12x – 10
c) h (x) = 2x² + 5x – 3	d) k (x) = x² – 4

a) f (x) = x² + 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) = –3x² + 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) = 2x² + 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) = x² – 4	i) vertex: (0, – 4) iii) minimum = – 4	ii) axis: x = 0, the y-axis iv) domain: R; v) range: .

a) f (x) = 2x² + 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) = –5x² + 4x + 1	0 = (1 + 5x)(1 – x)	x = 1 or x = – 1/5

d) f (x) = 2(x² – 8x + 17)	b² – 4ac = – 4	no Real solutions