1. write a quartic function with x-intercepts at - 5, 3, - 2, and - 4 with a y-intercept of 120.
2. write a quartic function with x-intercepts at - 4, 5, 7, and - 1 with a y-intercept of 280.
3. write a polynomial function with x-intercepts at - 2, $\frac{1}{2}$, $-\frac{3}{4}$, and 8 with a y-intercept of (0, 12).
4. convert $f(x) = 5(x - 4)(x + 6)(x - 2)$ to standard form.
5. use desmos to determine the x-intercepts of $y = 2x^3 + 4x^2 - 70x$. then, write the function in factored form.
6. use desmos to sketch a graph of $f(x) = (x - 3)(x + 6)(2x - 5)$. label the x-intercepts, y-intercept, local maximum, and local minimum.

Question

Accepted Answer

### Problem 1:
# Explanation:
## Step1: Start with factored form
A quartic function with x - intercepts $x=-5, x = 3, x=-2, x=-4$ can be written in factored form as $f(x)=a(x + 5)(x - 3)(x + 2)(x + 4)$, where $a$ is a constant.

## Step2: Find the value of $a$ using y - intercept
The y - intercept is the value of the function when $x = 0$. We know that the y - intercept is 120. Substitute $x = 0$ into the function:
$f(0)=a(0 + 5)(0 - 3)(0 + 2)(0 + 4)=a\times5\times(-3)\times2\times4=a\times(-120)$
Since $f(0) = 120$, we have $-120a=120$. Solving for $a$, we divide both sides by $-120$: $a=- 1$

## Step3: Write the function
Substitute $a=-1$ into the factored form:
$f(x)=-1(x + 5)(x - 3)(x + 2)(x + 4)$
We can also expand this, but the factored form with the correct $a$ is the function. If we expand:
First, multiply $(x + 5)(x - 3)=x^{2}+2x-15$ and $(x + 2)(x + 4)=x^{2}+6x + 8$
Then $(x^{2}+2x-15)(x^{2}+6x + 8)=x^{4}+6x^{3}+8x^{2}+2x^{3}+12x^{2}+16x-15x^{2}-90x - 120=x^{4}+8x^{3}+5x^{2}-74x-120$
Multiply by $-1$: $f(x)=-x^{4}-8x^{3}-5x^{2}+74x + 120$

# Answer: $f(x)=-(x + 5)(x - 3)(x + 2)(x + 4)$ (or $f(x)=-x^{4}-8x^{3}-5x^{2}+74x + 120$)

### Problem 2:
# Explanation:
## Step1: Write the factored form
A quartic function with x - intercepts $x=-4, x = 5, x = 7, x=-1$ is $f(x)=a(x + 4)(x - 5)(x - 7)(x + 1)$

## Step2: Find $a$ using y - intercept
The y - intercept is 280, so when $x = 0$, $f(0)=280$
$f(0)=a(0 + 4)(0 - 5)(0 - 7)(0 + 1)=a\times4\times(-5)\times(-7)\times1=a\times140$
Set $140a = 280$, then $a = 2$

## Step3: Write the function
Substitute $a = 2$ into the factored form: $f(x)=2(x + 4)(x - 5)(x - 7)(x + 1)$
If we expand:
First, $(x + 4)(x + 1)=x^{2}+5x + 4$ and $(x - 5)(x - 7)=x^{2}-12x + 35$
Then $(x^{2}+5x + 4)(x^{2}-12x + 35)=x^{4}-12x^{3}+35x^{2}+5x^{3}-60x^{2}+175x+4x^{2}-48x + 140=x^{4}-7x^{3}-21x^{2}+127x + 140$
Multiply by 2: $f(x)=2x^{4}-14x^{3}-42x^{2}+254x + 280$

# Answer: $f(x)=2(x + 4)(x - 5)(x - 7)(x + 1)$ (or $f(x)=2x^{4}-14x^{3}-42x^{2}+254x + 280$)

### Problem 3:
# Explanation:
## Step1: Write the factored form
A polynomial function with x - intercepts $x=-2, x=\frac{1}{2}, x =-\frac{3}{4}, x = 8$ can be written as $f(x)=a(x + 2)(x-\frac{1}{2})(x+\frac{3}{4})(x - 8)$
To eliminate fractions, we can rewrite the factors with denominators: $x-\frac{1}{2}=\frac{2x - 1}{2}$ and $x+\frac{3}{4}=\frac{4x+3}{4}$
So $f(x)=a(x + 2)\times\frac{2x - 1}{2}\times\frac{4x + 3}{4}\times(x - 8)=a\times\frac{(x + 2)(2x - 1)(4x + 3)(x - 8)}{8}$

## Step2: Find $a$ using y - intercept
The y - intercept is 12 (when $x = 0$)
$f(0)=a\times\frac{(0 + 2)(2\times0 - 1)(4\times0+3)(0 - 8)}{8}=a\times\frac{2\times(-1)\times3\times(-8)}{8}=a\times\frac{48}{8}=6a$
Set $6a = 12$, so $a = 2$

## Step3: Write the function
Substitute $a = 2$ into the function:
$f(x)=2\times\frac{(x + 2)(2x - 1)(4x + 3)(x - 8)}{8}=\frac{(x + 2)(2x - 1)(4x + 3)(x - 8)}{4}$
If we expand:
First, multiply $(x + 2)(x - 8)=x^{2}-6x - 16$ and $(2x - 1)(4x + 3)=8x^{2}+6x-4x - 3=8x^{2}+2x - 3$
Then $(x^{2}-6x - 16)(8x^{2}+2x - 3)=8x^{4}+2x^{3}-3x^{2}-48x^{3}-12x^{2}+18x-128x^{2}-32x + 48=8x^{4}-46x^{3}-143x^{2}-14x + 48$
Divide by 4: $f(x)=2x^{4}-\frac{23}{2}x^{3}-\frac{143}{4}x^{2}-\frac{7}{2}x + 12$ (or keep in factored form: $f(x)=\frac{1}{2}(x + 2)(2x - 1)(4x + 3)(x - 8)$ after correcting the $a$ calculation. Wait, let's re - calculate $f(0)$ again.
Original factored form: $f(x)=a(x + 2)(x-\frac{1}{2})(x+\frac{3}{4})(x - 8)$
$f(0)=a(2)(-\frac{1}{2})(\frac{3}{4})(-8)=a\times(2\times(-\frac{1}{2})\times\frac{3}{4}\times(-8))=a\times( - 1\times\frac{3}{4}\times(-8))=a\times6$
We know $f(0) = 12$, so $6a=12$, $a = 2$
So $f(x)=2(x + 2)(x-\frac{1}{2})(x+\frac{3}{4})(x - 8)$
Multiply $(x-\frac{1}{2})$ by 2 and $(x+\frac{3}{4})$ by 4 to eliminate fractions:
$f(x)=2\times\frac{2(x-\frac{1}{2})}{2}\times\frac{4(x+\frac{3}{4})}{4}\times(x + 2)(x - 8)=2\times\frac{(2x - 1)}{2}\times\frac{(4x + 3)}{4}\times(x + 2)(x - 8)=\frac{(2x - 1)(4x + 3)(x + 2)(x - 8)}{2}$
Now, $f(0)=\frac{(-1)(3)(2)(-8)}{2}=\frac{48}{2}=24$. Wait, I made a mistake earlier. The y - intercept is $(0,12)$, so $f(0) = 12$
So $\frac{(2x - 1)(4x + 3)(x + 2)(x - 8)}{2}$ at $x = 0$ is $\frac{(-1)(3)(2)(-8)}{2}=\frac{48}{2}=24$. So we need to adjust $a$. Let $f(x)=a(x + 2)(x-\frac{1}{2})(x+\frac{3}{4})(x - 8)$
$f(0)=a(2)(-\frac{1}{2})(\frac{3}{4})(-8)=a\times(2\times(-\frac{1}{2})=-1; - 1\times\frac{3}{4}=-\frac{3}{4};-\frac{3}{4}\times(-8) = 6)$, so $f(0)=6a$. We need $6a = 12$, so $a = 2$ was wrong. Wait, $6a=12$ implies $a = 2$, but when we plug $a = 2$ into $f(x)=2(x + 2)(x-\frac{1}{2})(x+\frac{3}{4})(x - 8)$, $f(0)=2\times6 = 12$, which is correct. Wait, my expansion was wrong. Let's recalculate $f(0)$ with $a = 2$:
$f(0)=2\times(2)\times(-\frac{1}{2})\times(\frac{3}{4})\times(-8)=2\times(2\times(-\frac{1}{2})=-1; - 1\times\frac{3}{4}=-\frac{3}{4};-\frac{3}{4}\times(-8)=6)=2\times6 = 12$, which is correct. So the earlier expansion mistake was mine. Let's expand $(x + 2)(x-\frac{1}{2})=x^{2}-\frac{1}{2}x+2x - 1=x^{2}+\frac{3}{2}x - 1$
$(x+\frac{3}{4})(x - 8)=x^{2}-8x+\frac{3}{4}x - 6=x^{2}-\frac{29}{4}x - 6$
Now multiply $(x^{2}+\frac{3}{2}x - 1)(x^{2}-\frac{29}{4}x - 6)$
$=x^{4}-\frac{29}{4}x^{3}-6x^{2}+\frac{3}{2}x^{3}-\frac{87}{8}x^{2}-9x - x^{2}+\frac{29}{4}x + 6$
$=x^{4}+(-\frac{29}{4}+\frac{6}{4})x^{3}+(-6-\frac{87}{8}-1)x^{2}+(-9+\frac{29}{4})x + 6$
$=x^{4}-\frac{23}{4}x^{3}-\frac{139}{8}x^{2}-\frac{7}{4}x + 6$
Multiply by 2: $f(x)=2x^{4}-\frac{23}{2}x^{3}-\frac{139}{4}x^{2}-\frac{7}{2}x + 12$

# Answer: $f(x)=2(x + 2)(x-\frac{1}{2})(x+\frac{3}{4})(x - 8)$ (or the expanded form $2x^{4}-\frac{23}{2}x^{3}-\frac{139}{4}x^{2}-\frac{7}{2}x + 12$)

### Problem 4:
# Explanation:
## Step1: Multiply two factors first
First, multiply $(x - 4)(x + 6)$:
$(x - 4)(x + 6)=x^{2}+6x-4x - 24=x^{2}+2x - 24$

## Step2: Multiply the result by $(x - 2)$
$(x^{2}+2x - 24)(x - 2)=x^{3}-2x^{2}+2x^{2}-4x-24x + 48=x^{3}-28x + 48$

## Step3: Multiply by 5
$f(x)=5(x^{3}-28x + 48)=5x^{3}-140x + 240$

# Answer: $f(x)=5x^{3}-140x + 240$

### Problem 5:
# Explanation:
## Step1: Factor out the greatest common factor
Given $y = 2x^{3}+4x^{2}-70x$, factor out $2x$:
$y=2x(x^{2}+2x - 35)$

## Step2: Factor the quadratic
Factor $x^{2}+2x - 35$. We need two numbers that multiply to $-35$ and add to $2$. The numbers are $7$ and $-5$. So $x^{2}+2x - 35=(x + 7)(x - 5)$

## Step3: Write the factored form
$y = 2x(x + 7)(x - 5)$

## Step4: Find x - intercepts
To find the x - intercepts, set $y = 0$:
$2x(x + 7)(x - 5)=0$
So $x = 0$ or $x=-7$ or $x = 5$

# Answer: The factored form is $y = 2x(x + 7)(x - 5)$ and the x - intercepts are $x = 0$, $x=-7$, $x = 5$

### Problem 6:
# Explanation:
## Step1: Find x - intercepts
To find the x - intercepts, set $f(x)=0$:
$(x - 3)(x + 6)(2x - 5)=0$
So $x - 3=0$ or $x + 6=0$ or $2x - 5=0$
Solving these, we get $x = 3$, $x=-6$, $x=\frac{5}{2}$

## Step2: Find y - intercept
To find the y - intercept, set $x = 0$:
$f(0)=(0 - 3)(0 + 6)(2\times0 - 5)=(-3)(6)(-5)=90$

## Step3: Find local maximum and minimum (using Desmos or calculus)
Using Desmos, when we graph $f(x)=(x - 3)(x + 6)(2x - 5)$, we can find the local maximum and minimum.
First, expand the function: $(x - 3)(x + 6)=x^{2}+3x - 18$; $(x^{2}+3x - 18)(2x - 5)=2x^{3}-5x^{2}+6x^{2}-15x-36x + 90=2x^{3}+x^{2}-51x + 90$
To find local extrema, take the derivative $f^\prime(x)=6x^{2}+2x - 51$
Set $f^\prime(x)=0$: $6x^{2}+2x - 51 = 0$
Using the quadratic formula $x=\frac{-2\pm\sqrt{4+1224}}{12}=\frac{-2\pm\sqrt{1228}}{12}=\frac{-2\pm2\sqrt{307}}{12}=\frac{-1\pm\sqrt{307}}{6}$
$\sqrt{307}\approx17.52$, so $x\approx\frac{-1 + 17.52}{6}\approx2.75$ and $x\approx\frac{-1-17.52}{6}\approx - 3.09$
Then find the corresponding $y$ -

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step1: Start with factored form

Step2: Find the value of \(a\) using y - intercept

Step3: Write the function

Step1: Write the factored form

Step2: Find \(a\) using y - intercept

Step3: Write the function

Step1: Write the factored form

Step2: Find \(a\) using y - intercept

Step3: Write the function

Answer:

Problem 2: