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lara [203]
3 years ago
5

given an existing function: f(x)=0.5(x-2)2+3, what transformstiins would have to be made to result in g(x)=-2(x+3)2 -1?

Mathematics
1 answer:
eimsori [14]3 years ago
5 0

Answer:

  • vertical scaling by a factor of -4
  • horizontal translation 5 units left
  • vertical translation 11 units up

Step-by-step explanation:

We notice that the multiplier of the squared term in f(x) is 0.5; in g(x), it is -2, so is a factor of -4 times that in f(x).

If we scale f(x) by a factor of -4, we get ...

  -4f(x) = -2(x -2)² -12

In order for the squared quantity to be x+3, we have to add 5 to the value that is squared in f(x). That is, x -2 must become x +3. We have to replace x with (x+5) to do that, so ...

  (x+5) -2 = x +3

The replacement of x with x+5 amounts to a translation of 5 units to the left.

We note that the added constant after our scaling changes from +3 to -12. Instead, we want it to be -1, so we must add 11 to the scaled function. That translates it upward by 11 units.

The attached graph shows the scaled and translated function g(x):

  g(x) = -4f(x +5) +11

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This should help you

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2 years ago
Can you answer these please many thanks
givi [52]

Given: (a) 2(x+3)=2x+6 and (b) 4(y-3)=4y-12.

To find: The expanded form of the given expressions.

(c) 4(m+n)=4m+4n       (using a(b+c)=ab+ac)

(d) 3(5-q)=15-3q           (using a(b-c)=ab-ac)

(e) 5(2c+1)=10c+5          (using a(b+c)=ab+ac)

(f) 3(2x-5)=6x-15          (using a(b-c)=ab-ac)

(g) 7(4b-1)=28b-7          (using a(b-c)=ab-ac)

(h) 3(2x+y-5)=3((2x+y)-5)

⇒3(2x+y-5)=3(2x+y)-15         (using a(b-c)=ab-ac)

⇒3(2x+y-5)=6x+3y-15            (using a(b+c)=ab+ac)

(i) 2(6a-4b+3)=2((6a-4b)+3)

⇒2(6a-4b+3)=2(6a-4b)+6         (using a(b+c)=ab+ac)

⇒2(6a-4b+3)=12a-8b+6            (using a(b-c)=ab-ac)

(j)6(m+n+p)=6((m+n)+p)

⇒ 6(m+n+p)=6(m+n)+6p           (using a(b+c)=ab+ac)

⇒ 6(m+n+p)=6m+6n+6p            (using a(b+c)=ab+ac)

(k) y(y+2)=y^2+2y          (using a(b+c)=ab+ac)

(l) g(g-3)=g^2-3g           (using a(b-c)=ab-ac)

(m) n(4-n)=4n-n^2        (using a(b-c)=ab-ac)

(n) a(b+c)=ab+ac           (using a(b+c)=ab+ac)

(o) s(3s-4)=3s^2-4s        (using a(b-c)=ab-ac)

(p) 2x(x+5)=2x^2+10x     (using a(b+c)=ab+ac)

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(t) 5p(4p-5q)=20p^2-25pq    (using a(b-c)=ab-ac)

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