How to estimately graph a quartic and quadratic function

[Chaim]

7(x-3)²(x+5)⁴

That means the zeroes would be 3, and -5
Then wouldn't the 7 have an exponent of 1? So it would be 7¹
Therefore the graph would go through the x-axis, then bounces off the x-axis after it hits the even multiplicity or even exponents?
So like this?

 

[lookagain]

7(x-3)²(x+5)⁴

That means the zeroes would be 3, and -5
Then wouldn't the 7 have an exponent of 1? So it would be 7¹
Therefore the graph would go through the x-axis, then bounces off the
x-axis after it hits the even multiplicity or even exponents?
View attachment 1472So like this?

Chaim,


the issue is what is the degree of 7? The degree of 7 is 0.

The degree of that polynomial is still 6.

 

[soroban]

Hello, Chaim!

\(\displaystyle y \:=\:7(x-3)^2(x+5)^4\)

That means the zeroes would be 3, and -5. . Right!
Then wouldn't the 7 have an exponent of 1? . So it would be 7¹ . Yes, but so what?

Therefore, the graph would go through the x-axis, . No
then bounces off the x-axis when it hits the even multiplicity or even exponents? . Yes!

The graph has x-intercepts of 3 and -5.
Since they have even multiplicity, the graph is tangent to the x-axis at those points.

The function is a sixth-degree polynomial with a positive leading coefficient.
. . Hence, it opens upward. . \(\displaystyle \begin{array}{c}\cap\;\cap \\ [-2.5mm]\cup\,\cup\,\cup \end{array}\)


The derivative is: .\(\displaystyle y' \:=\:14(x-3)(x+5)^3(3x-1)\)

The critical values are: .\(\displaystyle \begin{Bmatrix} x = \text{-}5 & \text{min} \\ x = \frac{1}{3} & \text{max} \\ x = 3 & \text{min} \end{Bmatrix}\)


Hence, the graph is shaped like a \(\displaystyle W.\)

                  |
      *           |               *
                  |
       *          | *            *
        *        *| :  *        *
          *     * | :   *     *
    ---------*----+-+------*---------
            -5    |1/3     3
                  |

 

[Chaim]

Chaim,


the issue is what is the degree of 7? The degree of 7 is 0.

The degree of that polynomial is still 6.

Oh I see thanks
I was overthinking myself, believing it was the degree of 7 xD
But I forgot it's actually the degree of 6 if you multiply all the exponents together

 

[Chaim]

Hello, Chaim!


The graph has x-intercepts of 3 and -5.
Since they have even multiplicity, the graph is tangent to the x-axis at those points.

The function is a sixth-degree polynomial with a positive leading coefficient.
. . Hence, it opens upward. . \(\displaystyle \begin{array}{c}\cap\;\cap \\ [-2.5mm]\cup\,\cup\,\cup \end{array}\)


The derivative is: .\(\displaystyle y' \:=\:14(x-3)(x+5)^3(3x-1)\)

The critical values are: .\(\displaystyle \begin{Bmatrix} x = \text{-}5 & \text{min} \\ x = \frac{1}{3} & \text{max} \\ x = 3 & \text{min} \end{Bmatrix}\)


Hence, the graph is shaped like a \(\displaystyle W.\)

                  |
      *           |               *
                  |
       *          | *            *
        *        *| :  *        *
          *     * | :   *     *
    ---------*----+-+------*---------
            -5    |1/3     3
                  |

Oh wow thanks!
I am starting to understand this a bit more now!
At first I thought it meant it will never pass through the x-axis if it was even, but I understand it can bounce off from the top or bottom xD
And I see where you got your maximum point now by factoring, then using the vertex formula (I believe so)

Just wondering, why does it shape like a 'w'?
Cause I thought 'w' were usually quartics, so I overthought again, and thought it would be some weird shape with 6 curves xD

 

[lookagain]

Oh I see thanks
I was overthinking myself, believing it was the degree of 7 xD
But I forgot it's actually the degree of 6 if you \(\displaystyle > > \)multiply \(\displaystyle < < \)all the exponents together

No, you \(\displaystyle \text{add}\) all of the (pertinent) exponents together.

So, 2 + 4 = 6.

Then, it is a 6th degree polynomial.