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Sladkaya [172]
3 years ago
15

Graph the function in the coordinate plane. Use the Mark Feature tool to indicate the x- and y-intercepts of the function.y=2/3x

+4

Mathematics
1 answer:
Darina [25.2K]3 years ago
6 0

Answer:

My blue dot is the y-intercept.

My red dot is my x-intercept.

Please look at the graph.

Step-by-step explanation:

I can show you my graph and mark it where the x-intercepts and y-intercepts are.

Let's begin.

We have y=2/3 x+4.

Compare this to the slope-intercept form, y=mx+b where m is the slope and b is the y-intercept.

You should see that m=2/3 and b=4.

This means the slope is 2/3 and the y-intercept is 4.

Don't forget slope means rise/run.

So once we graph 4 (plot a point) on the y-axis, then we will use our slope to get to one more point.  The slope here tells us to rise 2 and run 3.

Now sometimes our graph is not accurate when drawing by hand so there is a way without graphing that you can find the x- and y-intercepts.

The x-intercept is when the y-coordinate is 0.

The y-intercept is when the x-coordinate is 0.

So to find the x-intercept, I'm going to set y to 0 and solve for x. Like so,

0=2/3 x +4

Subtract 4 on both sides:

-4=2/3 x

Multiply both sides by the reciprocal of 2/3 which is 3/2:

3/2 (-4)=x

Simplify:

-12/2=x

Simplify:

-6=x

Symmetric Property:

x=-6

So the x-intercept is (-6,0).

I actually already have the y-intercept since my equation is in y=mx+b (slope-intercept form).  But if it wasn't you could just set x to 0 and solve for y. Like so:

y=2/3 (0)+4

y=0+4

y=4

The y-intercept is (0,4).

Let's go to our graph now.

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Answer:

\displaystyle J'(3) = -1

General Formulas and Concepts:

<u>Algebra I</u>

  • Functions
  • Function Notation

<u>Calculus</u>

Derivatives

Derivative Notation

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

Derivative:                                                                                                                     \displaystyle \frac{d}{dx} [e^u]=e^u \cdot u'

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em />\displaystyle J(x) = e^{f(x)}<em />

<em />

<u>Step 2: Differentiate</u>

  1. eˣ Derivative [Derivative Rule - Chain Rule]:                                                  \displaystyle J'(x) = \frac{d}{dx}[e^{f(x)}] \cdot \frac{d}{dx}[f(x)]
  2. Simplify:                                                                                                             \displaystyle J'(x) = f'(x)e^{f(x)}

<u>Step 3: Evaluate</u>

  1. Substitute in <em>x</em> [Derivative]:                                                                              \displaystyle J'(3) = f'(3)e^{f(3)}
  2. Substitute in function values:                                                                          \displaystyle J'(3) = -e^{0}
  3. Simplify:                                                                                                             \displaystyle J'(3) = -1

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

Unit: Derivatives

Book: College Calculus 10e

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