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Gnom [1K]
3 years ago
13

Use the parabola tool to graph the quadratic function f(x)=(x−4)(x+2). Graph the parabola by first plotting its vertex and then

plotting a second point on the parabola.

Mathematics
2 answers:
mote1985 [20]3 years ago
4 0

Here are a bunch of CORRECT answers, your answer is somewhere in there. For the first CORRECT answer the second point is -5,-9. Don't make the same mistake I did on question 3, but it still shows the correct answer. I love to help.

posledela3 years ago
3 0
If the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

Then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

One way in which you can find the vertex is to calculate x = -b / (2a).  Here, that comes to
     
        -(-2)
x = ---------- = 1.  f(1) = 1^2 - 2(1) - 8 = - 9.  Vertex is at (1, -9).
        2(1) 

Find another point on the curve:  Let x=0, for which y=-8.  Then (0, -8) is on the curve.  Plot the vertex and this point and draw a smooth parabola thru both.


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5 0
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Sholpan [36]
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4 0
3 years ago
Please someone help me with the question
Yuki888 [10]

Answer:

\displaystyle \frac{d}{dx}[e^{2x}] = 2e^{2x}

\displaystyle \frac{d}{dx}[e^{3x}] = 3e^{3x}

General Formulas and Concepts:

<u>Algebra I</u>

  • Terms/Coefficients
  • Exponential Rule [Multiplying]:                                                                      \displaystyle b^m \cdot b^n = b^{m + n}

<u>Calculus</u>

Derivatives

Derivative Notation

eˣ Derivative:                                                                                                           \displaystyle \frac{d}{dx}[e^x] = e^x

Derivative Rule [Product Rule]:                                                                                  \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle \frac{d}{dx}[e^{2x}] = \frac{d}{dx}[e^x \cdot e^x]<u />

<u />\displaystyle \frac{d}{dx}[e^{3x}] = \frac{d}{dx}[e^x \cdot e^{2x}]<u />

<u />

<u>Step 2: Differentiate</u>

<u />\displaystyle \frac{d}{dx}[e^{2x}]<u />

  1. [Derivative] Product Rule:                                                                              \displaystyle \frac{d}{dx}[e^{2x}] = \frac{d}{dx}[e^x]e^x + e^x\frac{d}{dx}[e^x]
  2. [Derivative] eˣ Derivative:                                                                               \displaystyle \frac{d}{dx}[e^{2x}] = e^x \cdot e^x + e^x \cdot e^x
  3. [Derivative] Multiply [Exponential Rule - Multiplying]:                                  \displaystyle \frac{d}{dx}[e^{2x}] = e^{2x} + e^{2x}
  4. [Derivative] Combine like terms [Addition]:                                                  \displaystyle \frac{d}{dx}[e^{2x}] = 2e^{2x}

\displaystyle \frac{d}{dx}[e^{3x}]

  1. [Derivative] Product Rule:                                                                              \displaystyle \frac{d}{dx}[e^{3x}] = \frac{d}{dx}[e^x]e^{2x} + e^x\frac{d}{dx}[e^{2x}]
  2. [Derivative] eˣ Derivatives:                                                                             \displaystyle \frac{d}{dx}[e^{3x}] = e^x(e^{2x}) + e^x(2e^{2x})
  3. [Derivative] Multiply [Exponential Rule - Multiplying]:                                  \displaystyle \frac{d}{dx}[e^{3x}] = e^{3x} + 2e^{3x}
  4. [Derivative] Combine like terms [Addition]:                                                  \displaystyle \frac{d}{dx}[e^{3x}] = 3e^{3x}

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

8 0
3 years ago
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