126k views
2 votes
Please someone help me out.

How many x-intercepts does the graph of
y = x³ - 1 have?
JO
K
1
L 2
M 3
N An infinite number

User Aldan
by
7.3k points

1 Answer

3 votes

Answer:

K. 1

Explanation:

You want to know the number of x-intercepts of the graph of y = x³ -1.

Descartes rule of signs

When the expression is examined left to right, there is one sign change of the coefficients. The sign of x³ is positive, and the sign of -1 is negative. This means there is one positive real x-intercept (root).

When the signs of the odd degree terms are reversed, we have ...

y = -x³ -1

Both terms have a negative sign, so there are 0 sign changes, meaning there are 0 negative real x-intercepts (roots).

Factors

The factoring of the difference of two cubes is ...

a³ -b³ = (a -b)(a² +ab +b²)

For a=x and b=1, the factoring is ...

y = x³ -1 = (x -1)(x² +x +1)

The discriminant of the quadratic factor is ...

d = 1² -4(1)(1) = -3

Since it is negative, we know the roots of this factor are complex. The only x-intercept is the one contributed by the factor (x -1), which is zero for x=1.

There is one (1) x-intercept on the graph of y = x³ -1, choice K.

__

Additional comment

The discriminant of the quadratic ax²+bx+c is d=b²-4ac. When it is negative, the quadratic has 2 complex roots.

<951414049393>

Please someone help me out. How many x-intercepts does the graph of y = x³ - 1 have-example-1
User Denispyr
by
8.1k points