Mathematics: Will give for correct answer. : ) consider the function f(x) = x^3 + 2x^2 -3 . (a) graph the function. (b) what are the x- and y-intercepts of the graph? i already know how this function looks on a graph. all i want to know is how to do it.

21.5 degreesSubtract 180 from 48.3 and then add it to 26.8. Take this value and subtract it from 180 and you will get a final answer of 21.5...

Will give for correct answer. : ) consider the function

MoltenSansriel1933

11.01.2022

(a) The graph is shown below.

(b)

x-intercept = (1, 0)

y-intercept = (0, -3)

Step-by-step explanation:

Given:

The function is given as:

f(x)=x^3+2x^2-3

(a)

In order to plot the graph, we need to find some points on it. We also need to find the maximum and minimum of the function.

Let us find the maximum and minimum of the function. The maxima or minima occurs when the derivative of the function is 0.

Differentiating the function with respect to 'x', we get:

f'(x)=3x^2+4x=x(3x+4)

Now, equating f'(x) to 0, we solve for 'x'. This gives,

$x(3x+4)=0\\x=0\ or\ 3x+4=0\\x=0\or\ x=-\frac{4}{3}$

Now, let us find the 'y' values for the above 'x' values.

When x=0,y=0+0-3=-3

When $x=-\frac{4}{3},y=(-\frac{4}{3})^3+2(-\frac{4}{3})^2-3=-1.8$

So, the point of maxima is (-1.33, -1.8) and point of minima is (0, -3). Mark these two points on the graph.

Let us take another random point for 'x'. Let x=-1

y=(-1)^3+2(-1)^2-3=-2 . Mark the point (-1, -2) on the graph.

Now, let x=1 . The value of 'y' is:

y=(1)^3+2(1)^2-3=0 . Mark the point (1, 0) on the graph.

Now, draw a smooth curve passing through all of these points with a maximum curve at (-1.33, -1.8) and minimum curve at (0, -3).

The graph is shown below.

(b)

The x-intercept is at the point when 'y' value is 0. So, the point on the graph is (1, 0).

The y-intercept is at the point when 'x' value is 0. So, the point on the graph is (0, -3).

0,0(0 оценок)