What transformations of the parent graph of f(x)=c produce the graphs of the following functions? a) m(x)=7x−3.5−10 b) j(x)=−212x+4

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Algebra
College algebra
Transformations of functions

Jaden Easton

Answered question

2021-08-14

What transformations of the parent graph of

f (x) = \sqrt{c}

produce the graphs of the following functions?
a)

m (x) = \sqrt{7 x - 3.5} - 10

j (x) = - 2 \sqrt{12 x} + 4

Answer & Explanation

okomgcae

Skilled2021-08-15Added 93 answers

a)First, we have to rewrite the function m(x) in the form $m (x) = \sqrt{x — h} + k$
So, the steps are: factor the radicand, use the Product Property of Radicals and than simplify, so, we have next:
$m (x) = \sqrt{7 (x - 0.5)} - 10$
$= \sqrt{7} \cdot \sqrt{x - 0.5} - 10$
$= 2.65 \sqrt{x} = 0.5 - 10$
So, the graph of m(x) is a vertical stretch of a parent function by a factor of 2.65 or $\sqrt{7}$ , translated right 0.5 units and translated down 10 units. On the following picture there is a graph of $y = \sqrt{x} and m (x) = \sqrt{7 x - 3.5} - 10$

b)First, we have to rewrite the function j(x) in the form: $j (x) = a \sqrt{x - h} + k$ So, the steps are: use the Product Property of Radicals and than simplify, so we have next:
$j (x) = - 2 \sqrt{4 \cdot 3 \cdot x} + 4$
$= - 2 \sqrt{4} \cdot \sqrt{3} \cdot \sqrt{x} + 4$
$= - 2 \cdot 2 \cdot \sqrt{3} \cdot \sqrt{x} + 4$
$= - 4 \sqrt{3} \cdot \sqrt{x} + 4$
So, the graph of j(x) is a vertical stretch of a parent function by a factor of $- 4 \sqrt{3}$ , translated up 4 units.

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