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 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) = 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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