Answer:
14) vertex: (3, -6); focus: (3, -5.5); directrix: y = -6.5
17) see below for a graph
19) (±3, ±1) — four points
20) see below for the work
Step-by-step explanation:
14) The vertex form of a quadratic equation can be written as ...
[tex]y=\dfrac{1}{4p}(x-h)^2+k[/tex]
This parabola will have its vertex at (h, k), its focus at (h, k+p), and its directrix at y=k-p.
The given equation is ...
y = 1/2(x -3)² -6
Comparing this to the form above, we see that ...
4p = 2, h = 3, k = -6
From this, we can find the value of p to be ...
p = 2/4 = 1/2 . . . . divide the above equation by 4
Now, we know the parameters of the parabola are ...
- vertex = (h, k) = (3, -6)
- focus = (h, k+p) = (3, -6 +1/2) = (3, -5.5)
- directrix ⇒ y = k-p = -6 -1/2 = -6.5
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17) The equation of the parabola is given as ...
y = 1/4(x -1)² +5
Comparing the given equation to the form shown in the above problem, we see that ...
- 4p = 4 ⇒ p = 1
- (h, k) = (1, 5)
So the parameters of the parabola are ...
- vertex = (1, 5)
- focus = (1, 5+1) = (1, 6)
- directrix ⇒ y = 5-1 = 4
The graph is shown below.
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20) It is convenient to show the solution steps before showing the solution. Hence we show problem 20 (steps) before problem 19 (solution).
The given system of equations is ...
The two equations can be added to give ...
(x² +4y²) +(x² -4y²) = (13) +(5)
2x² = 18 . . . . . . eliminate parentheses, collect terms
x² = 9 . . . . . . . . divide by 2
x = ±3 . . . . . . . . take the square root
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The second equation can be subtracted from the first to give ...
(x² +4y²) -(x² -4y²) = (13) -(5)
8y² = 8 . . . . . . . eliminate parentheses, collect terms
y² = 1 . . . . . . . . . divide by 8
y = ±1 . . . . . . . . . take the square root
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19) Based on the above, the four solutions to the given system of equations are ...
(3, 1), (3, -1), (-3, 1), (-3, -1)
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A graph is shown in the second attachment.
