Answer:
(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:
[tex]f(x)=x^3+2x^2-3[/tex]
(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:
[tex]f'(x)=3x^2+4x=x(3x+4)[/tex]
Now, equating [tex]f'(x)[/tex] to 0, we solve for 'x'. This gives,
[tex]x(3x+4)=0\\x=0\ or\ 3x+4=0\\x=0\or\ x=-\frac{4}{3}[/tex]
Now, let us find the 'y' values for the above 'x' values.
When [tex]x=0,y=0+0-3=-3[/tex]
When [tex]x=-\frac{4}{3},y=(-\frac{4}{3})^3+2(-\frac{4}{3})^2-3=-1.8[/tex]
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 [tex]x=-1[/tex]
[tex]y=(-1)^3+2(-1)^2-3=-2[/tex]. Mark the point (-1, -2) on the graph.
Now, let [tex]x=1[/tex]. The value of 'y' is:
[tex]y=(1)^3+2(1)^2-3=0[/tex]. 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).
