Question

Consider the function
f (x) = x^2-3.

Answer 1

Your graph falls to the left and raises to the right

Your X intercepts are

-2, -1, 0, 1, 2

In increasing order
Your Y intercepts are

-3, -2, -3, 0,

13 in increasing order

Graph is included below

Answer 2

(a) See attached graph.

(b) x-intercept: (1, 0)
y-intercept: (0, -3)

Explanation:

The given function f(x) = x³ + 2x² - 3 is a cubic function with a positive leading coefficient. This means that the end behavior of the graph is:

• As x → -∞, f(x) → -∞
• As x → +∞, f(x) → +∞

So, the graph will start in the quadrant III and end in the quadrant I of the Cartesian plane.

As f(1) = 0, we know that (x - 1) is a factor of the function. Therefore:

`f(x)=(x-1)(x^2+3x+3)`

To find the x-intercepts, set the factors to zero and solve for x:

`\begin{aligned}x-1&=0\\x-1+1&=0+1\\x&=1\end{aligned}`

`\begin{aligned}x^2+3x+3&=0\\\\x^2+3x&=-3\\\\x^2+3x+(9)/(4)&=-3+(9)/(4)\\\\\left(x+(3)/(2)\right)^2&=-(3)/(4)\\\\x+(3)/(2)&=\pm(√(3))/(2)\:i\\\\x&=-(3)/(2)\pm(√(3))/(2)\:i\\\\x&=(-3\pm √(3)\:i)/(2)\end{aligned}`

Therefore, there is only one x-intercept at (1, 0).

To find the y-intercept, substitute x = 0 into the function:

`f(0) = (0)^3 + 2(0)^2 - 3`

`f(0)=0+0-3`

`f(0)=-3`

Therefore, the y-intercept is at (0, -3).

To help graph the function, find the coordinates of the local minimum and local maximum. To do this, differentiate the function:

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

Factor:

`f'(x)=x(3x+4)`

Now, set the first derivative to zero and solve for x to find the x-coordinates of the turning points of the curve:

`\begin{aligned}x(3x+4)&=0\\\\x&=0\implies x=0\\3x+4&=0 \implies x=-(4)/(3)\end{aligned}`

Therefore, the x-coordinates of the turning points are x = 0 and x = -⁴/₃.

We already know that y = -3 when x = 0. Therefore, the local minimum is also the y-intercept.

To find the y-coordinate of the local maximum, substitute x = -⁴/₃ into the function:

`f\left(-(4)/(3)\right) = \left(-(4)/(3)\right)^3 + 2\left(-(4)/(3)\right)^2 - 3`

`f\left(-(4)/(3)\right) = -(49)/(27)`

`f\left(-(4)/(3)\right) \approx -1.8`

Therefore, the local maximum is approximately (-1.3, -1.8).

To help graph the curve, find other points on the curve by substituting different values of x into the function:

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

`f(-1) = (-1)^3 + 2(-1)^2 - 3=-2`

`f(2) = (2)^3 + 2(2)^2 - 3=13`

To graph the function:

• Plot points (-2, -3), (-1.3, -1.8), (-1, -2), (0, -3), (1, 0) and (2, 13).
• Draw a smooth curve through the points that begins in quadrant III and ends in quadrant I. Ensure the curve changes direction at (-1.3, -1.8) and (0, -3).