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3 years ago

I need to find the domain and ranges of some of these graphs and I can’t really do the last few! Can anyone help?

Mathematics

1 answer:

hodyreva [135]3 years ago

3 0

(1) For the parabola on the bottom row, the domain would be R and the range would be y ≥ -5

(2) For the hyperbola on the bottom row, the domain would be R\{3} (since there is an asymptote at x = 3) and the range would be R\{4} (since there is an asymptote at y = 4)

(3) For the square root function on the bottom row, the domain would be x ≥ -5 and the range would be (-∞, -2]

(4) For the function to the very right on the bottom row, the domain would be R and the range would be (-∞, -3]

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Extended Problems, Extraneous Solutions

diamong [38]

Answer:

1a. 2 is the solution

1b. -3 is because square root of 1 doesn't equal -1.

1c.-3

Step-by-step explanation:

$(x + 2 = \sqrt{3x + 10}$

Square both sides

$(x + 2) {}^{2} = 3x + 10$

Remeber perfect square Rule

$(x + y) {}^{2} = {x}^{2} + 2xy + {y}^{2}$

So now we got

$x {}^{2} + 4x + 4 = 3x + 10$

${x}^{2} + x + 4 = 10$

${x}^{2} + x - 6 = 0$

Factor into binomial.

$(x + 3)(x - 2) = 0$

So our solution is 3 or 2. However, let's check our work.

$- 3 + 2 = \sqrt{3( - 3) + 10}$

$- 1 = \sqrt{1}$

However square root of 1 is 1. It could be 1 but in Math, We only take the principal square root which is positive always. So -3 isn't a solution.