Math / Algebra

QuestionThe graph of a polynomial function $Q(x)$ has a double root at $x=-2$ , a single root at $x=-6$ , and a double root at $x=3 x=3$ . $Q(x)$ also passes through the point $R(1,4)$ . - Determine the equation of $Q(x)$ and write your final answer in factored form. - What is the value of the $y$ -intercept for $Q(x)$ ?

Studdy Solution

STEP 1

What is this asking? We need to find the equation of a polynomial $Q(x)$ given its roots and a point it passes through, and then find its y-intercept. Watch out! Remember that "double root" means the root appears twice in the factored form.
Don't forget that there's a leading coefficient we need to find using the given point!

STEP 2

1. Set up the factored form.
2. Find the leading coefficient.
3. Write the full equation.
4. Calculate the y-intercept.

STEP 3

Alright, so we know that $Q(x)$ has a double root at $x = -2$ .
This means the factor $(x - (-2))$ or $(x+2)$ appears twice in the factored form.
So we have $(x+2)^2$ .

STEP 4

Next, we have a single root at $x = -6$ , which gives us the factor $(x - (-6))$ or $(x+6)$ .

STEP 5

And finally, another double root at $x = 3$ , giving us the factor $(x-3)^2$ .

STEP 6

Putting it all together, our factored form looks like this so far: $Q(x) = a(x+2)^2(x+6)(x-3)^2$ , where $a$ is the leading coefficient that we still need to find!

STEP 7

We know that $Q(x)$ passes through the point $R(1,4)$ .
This means when $x = 1$ , $Q(x) = 4$ .
Let's plug these values into our equation: $4 = a(1+2)^2(1+6)(1-3)^2$

STEP 8

Now, let's simplify: $4 = a(3)^2(7)(-2)^2$ $4 = a \cdot 9 \cdot 7 \cdot 4$ $4 = a \cdot 252$

STEP 9

To find $a$ , we divide both sides by 252: $a = \frac{4}{252}$ $a = \frac{1}{63}$

STEP 10

Now that we have our leading coefficient $a = \frac{1}{63}$ , we can write the complete factored form of $Q(x)$ : $Q(x) = \frac{1}{63}(x+2)^2(x+6)(x-3)^2$

STEP 11

The y-intercept occurs when $x = 0$ .
Let's plug in $x = 0$ into our equation: $Q(0) = \frac{1}{63}(0+2)^2(0+6)(0-3)^2$

STEP 12

Simplifying: $Q(0) = \frac{1}{63}(2)^2(6)(-3)^2$ $Q(0) = \frac{1}{63} \cdot 4 \cdot 6 \cdot 9$ $Q(0) = \frac{216}{63}$ $Q(0) = \frac{24}{7}$

STEP 13

The equation of $Q(x)$ in factored form is $Q(x) = \frac{1}{63}(x+2)^2(x+6)(x-3)^2$ .
The y-intercept is $\frac{24}{7}$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the factored form.
2. Find the leading coefficient.
3. Write the full equation.
4. Calculate the y-intercept.

STEP 3

Alright, so we know that $Q(x)$ has a double root at $x = -2$ .
This means the factor $(x - (-2))$ or $(x+2)$ appears twice in the factored form.
So we have $(x+2)^2$ .

STEP 4

Next, we have a single root at $x = -6$ , which gives us the factor $(x - (-6))$ or $(x+6)$ .

STEP 5

And finally, another double root at $x = 3$ , giving us the factor $(x-3)^2$ .

STEP 6

Putting it all together, our factored form looks like this so far: $Q(x) = a(x+2)^2(x+6)(x-3)^2$ , where $a$ is the leading coefficient that we still need to find!

STEP 7

We know that $Q(x)$ passes through the point $R(1,4)$ .
This means when $x = 1$ , $Q(x) = 4$ .
Let's plug these values into our equation: $4 = a(1+2)^2(1+6)(1-3)^2$

STEP 8

Now, let's simplify: $4 = a(3)^2(7)(-2)^2$ $4 = a \cdot 9 \cdot 7 \cdot 4$ $4 = a \cdot 252$

STEP 9

To find $a$ , we divide both sides by 252: $a = \frac{4}{252}$ $a = \frac{1}{63}$

STEP 10

Now that we have our leading coefficient $a = \frac{1}{63}$ , we can write the complete factored form of $Q(x)$ : $Q(x) = \frac{1}{63}(x+2)^2(x+6)(x-3)^2$

STEP 11

The y-intercept occurs when $x = 0$ .
Let's plug in $x = 0$ into our equation: $Q(0) = \frac{1}{63}(0+2)^2(0+6)(0-3)^2$

STEP 12

Simplifying: $Q(0) = \frac{1}{63}(2)^2(6)(-3)^2$ $Q(0) = \frac{1}{63} \cdot 4 \cdot 6 \cdot 9$ $Q(0) = \frac{216}{63}$ $Q(0) = \frac{24}{7}$

STEP 13

The equation of $Q(x)$ in factored form is $Q(x) = \frac{1}{63}(x+2)^2(x+6)(x-3)^2$ .
The y-intercept is $\frac{24}{7}$ .

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Studdy Solution

STEP 1

STEP 2

1. Set up the factored form.2. Find the leading coefficient.3. Write the full equation.4. Calculate the y-intercept.

STEP 3

Alright, so we know that Q(x)Q(x)Q(x) has a **double root** at x=−2x = -2x=−2.This means the factor (x−(−2))(x - (-2))(x−(−2)) or (x+2)(x+2)(x+2) appears **twice** in the factored form.So we have (x+2)2(x+2)^2(x+2)2.

STEP 4

Next, we have a **single root** at x=−6x = -6x=−6, which gives us the factor (x−(−6))(x - (-6))(x−(−6)) or (x+6)(x+6)(x+6).

STEP 5

And finally, another **double root** at x=3x = 3x=3, giving us the factor (x−3)2(x-3)^2(x−3)2.

STEP 6

Putting it all together, our factored form looks like this so far: Q(x)=a(x+2)2(x+6)(x−3)2Q(x) = a(x+2)^2(x+6)(x-3)^2Q(x)=a(x+2)2(x+6)(x−3)2, where aaa is the **leading coefficient** that we still need to find!

STEP 7

We know that Q(x)Q(x)Q(x) passes through the point R(1,4)R(1,4)R(1,4).This means when x=1x = 1x=1, Q(x)=4Q(x) = 4Q(x)=4.Let's plug these values into our equation: 4=a(1+2)2(1+6)(1−3)24 = a(1+2)^2(1+6)(1-3)^24=a(1+2)2(1+6)(1−3)2

STEP 8

Now, let's simplify: 4=a(3)2(7)(−2)24 = a(3)^2(7)(-2)^24=a(3)2(7)(−2)2 4=a⋅9⋅7⋅44 = a \cdot 9 \cdot 7 \cdot 44=a⋅9⋅7⋅44=a⋅2524 = a \cdot 2524=a⋅252

STEP 9

To find aaa, we **divide both sides** by 252: a=4252a = \frac{4}{252}a=2524​ a=163a = \frac{1}{63}a=631​

STEP 10

Now that we have our **leading coefficient** a=163a = \frac{1}{63}a=631​, we can write the complete factored form of Q(x)Q(x)Q(x): Q(x)=163(x+2)2(x+6)(x−3)2Q(x) = \frac{1}{63}(x+2)^2(x+6)(x-3)^2Q(x)=631​(x+2)2(x+6)(x−3)2

STEP 11

The y-intercept occurs when x=0x = 0x=0.Let's plug in x=0x = 0x=0 into our equation: Q(0)=163(0+2)2(0+6)(0−3)2Q(0) = \frac{1}{63}(0+2)^2(0+6)(0-3)^2Q(0)=631​(0+2)2(0+6)(0−3)2

STEP 12

Simplifying: Q(0)=163(2)2(6)(−3)2Q(0) = \frac{1}{63}(2)^2(6)(-3)^2Q(0)=631​(2)2(6)(−3)2 Q(0)=163⋅4⋅6⋅9Q(0) = \frac{1}{63} \cdot 4 \cdot 6 \cdot 9Q(0)=631​⋅4⋅6⋅9Q(0)=21663Q(0) = \frac{216}{63}Q(0)=63216​Q(0)=247Q(0) = \frac{24}{7}Q(0)=724​

STEP 13

The equation of Q(x)Q(x)Q(x) in factored form is Q(x)=163(x+2)2(x+6)(x−3)2Q(x) = \frac{1}{63}(x+2)^2(x+6)(x-3)^2Q(x)=631​(x+2)2(x+6)(x−3)2.The y-intercept is 247\frac{24}{7}724​.

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1. Set up the factored form.
2. Find the leading coefficient.
3. Write the full equation.
4. Calculate the y-intercept.

Alright, so we know that $Q(x)$ has a double root at $x = -2$ .
This means the factor $(x - (-2))$ or $(x+2)$ appears twice in the factored form.
So we have $(x+2)^2$ .

Next, we have a single root at $x = -6$ , which gives us the factor $(x - (-6))$ or $(x+6)$ .

And finally, another double root at $x = 3$ , giving us the factor $(x-3)^2$ .

Putting it all together, our factored form looks like this so far: $Q(x) = a(x+2)^2(x+6)(x-3)^2$ , where $a$ is the leading coefficient that we still need to find!

We know that $Q(x)$ passes through the point $R(1,4)$ .
This means when $x = 1$ , $Q(x) = 4$ .
Let's plug these values into our equation: $4 = a(1+2)^2(1+6)(1-3)^2$

Now, let's simplify: $4 = a(3)^2(7)(-2)^2$ $4 = a \cdot 9 \cdot 7 \cdot 4$ $4 = a \cdot 252$

To find $a$ , we divide both sides by 252: $a = \frac{4}{252}$ $a = \frac{1}{63}$

Now that we have our leading coefficient $a = \frac{1}{63}$ , we can write the complete factored form of $Q(x)$ : $Q(x) = \frac{1}{63}(x+2)^2(x+6)(x-3)^2$

The y-intercept occurs when $x = 0$ .
Let's plug in $x = 0$ into our equation: $Q(0) = \frac{1}{63}(0+2)^2(0+6)(0-3)^2$

Simplifying: $Q(0) = \frac{1}{63}(2)^2(6)(-3)^2$ $Q(0) = \frac{1}{63} \cdot 4 \cdot 6 \cdot 9$ $Q(0) = \frac{216}{63}$ $Q(0) = \frac{24}{7}$

The equation of $Q(x)$ in factored form is $Q(x) = \frac{1}{63}(x+2)^2(x+6)(x-3)^2$ .
The y-intercept is $\frac{24}{7}$ .