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)
.
2) Find Rule of Correspondence
(solution)
.
3) Find Zeros
(solution)
.
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: . |
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 – x_{1})(x – x_{2})
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) = 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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