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3 years ago

30 POINTS! The function P(x) = 3x2 + 4x + 5,is dilated by the function I(x) = P(2x). Write the new function I(x).

Mathematics

2 answers:

deff fn [24]3 years ago

5 0

Hey there!!!

Equation given :

p ( x ) = 3x² + 4x + 5

Find p ( 2x )

In order to solve this, we will have to substitute the value 2x instead of x

Hence,

... p ( 2x ) = 3 ( 2x )² + 4 ( 2x ) + 5

... p ( 2x ) = 3 ( 8x² ) + 8x + 5

... p ( 2x ) = 24x² + 8x + 5

As we know I ( x ) = p ( 2x )

The value of I(x) = 24x² + 8x + 5

Hope my answer helps!

kenny6666 [7]3 years ago

5 0

Answer:

$P(2x)=12x^{2} +8x+5$

Step-by-step explanation:

To obtain the new function indicated by I (x) = P (2x), it means that we must substitute in the original equation every time we find "x", the value of "2x

So, for $P(x)=3x^{2}+4x+5$

replacing I(x)=P(2x)

$P(2x)=3(2x)^{2} +4(2x)+5$

performing indicated operations

$P(2x)=3(4x^{2} )+8x+5$

$P(2x)=12x^{2} +8x+5$

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First, we are going to find the vertex of our quadratic. Remember that to find the vertex $(h,k)$ of a quadratic equation of the form $y=a x^{2} +bx+c$ , we use the vertex formula $h= \frac{-b}{2a}$ , and then, we evaluate our equation at $h$ to find $k$ .

We now from our quadratic that $a=2$ and $b=-32$ , so lets use our formula:
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$h= \frac{32}{4}$
$h=8$
Now we can evaluate our quadratic at 8 to find $k$ :
$k=2(8)^2-32(8)+56$
$k=2(64)-256+56$
$k=128-200$
$k=-72$
So the vertex of our function is (8,-72)

Next, we are going to use the vertex to rewrite our quadratic equation:
$y=a(x-h)^2+k$
$y=2(x-8)^2+(-72)$
$y=2(x-8)^2-72$
The x-coordinate of the minimum will be the x-coordinate of the vertex; in other words: 8.

We can conclude that:
The rewritten equation is $y=2(x-8)^2-72$
The x-coordinate of the minimum is 8