Determine the x-intercepts for
y = x^3 + 4x^2 - 4x - 16
and for
y = 2x^4 - 5x^2
Factor the expression. The intercepts are all rational roots.
Since the constant term is 16, and the coefficient of x³ is +1, the only rational factors to try are powers of 2.
You will find one intercept in the negative x region, and two in the positive x region.
The trial roots can be tested rather simply by evaluation the expression
y = f(x) = x^3 + 4x^2 - 4x - 16
For x=2,
f(2) = 2³ + 4(2)² -4(2) -16
= 16 + 8 - 8 - 16
=0
Therefore (x-2) is a factor, or x=+2 is an intercept.
Continue this way to find all the roots, or alternatively, reduce the expression by long division with (x-2).
Post your answer for a check if you wish.
For
y = 2x^4 - 5x^2
you can factor out the common factor x² to get
y=x²(2x²-5)
from which you can figure out the intercepts.
To determine the x-intercepts for the given equations, we need to find the values of x where the y-coordinate is zero. In other words, we are looking for the values of x that make the equations equal to zero.
Let's start with the first equation:
y = x^3 + 4x^2 - 4x - 16
To find the x-intercepts, we set y equal to zero:
0 = x^3 + 4x^2 - 4x - 16
Now, we can attempt to solve this equation for x. However, finding the exact solutions could be difficult in this case, as it is a cubic equation. Instead, we can use numerical methods or factoring to approximate the solutions.
One way to find the approximate x-intercepts is by factoring the equation. If we can factor it into linear and quadratic factors, we can solve for x more easily.
By testing different values of x, we find that x = -2 is a solution. Therefore, (x + 2) is a factor of the equation. We can perform synthetic division to divide the original equation by (x + 2):
-2 | 1 4 -4 -16
| -2 -4 16
--------------------
1 2 -8 0
The result of the division is the quadratic equation: x^2 + 2x - 8. Now, we can factor the quadratic equation:
0 = (x + 2)(x^2 + 2x - 8)
Setting each factor equal to zero, we have:
x + 2 = 0 => x = -2
x^2 + 2x - 8 = 0
To solve the quadratic equation, we can use factoring, completing the square, or the quadratic formula. In this case, factoring doesn't work easily, so we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For the quadratic equation x^2 + 2x - 8 = 0:
a = 1, b = 2, c = -8
Substituting the values into the formula:
x = (-2 ± √(2^2 - 4*1*(-8))) / (2*1)
Simplifying:
x = (-2 ± √(4 + 32)) / 2
x = (-2 ± √36) / 2
x = (-2 ± 6) / 2
This gives us two solutions:
x = (-2 + 6) / 2 = 4/2 = 2
x = (-2 - 6) / 2 = -8/2 = -4
Therefore, the x-intercepts for the equation y = x^3 + 4x^2 - 4x - 16 are x = -2, x = -4, and x = 2.
Now, let's move on to the second equation:
y = 2x^4 - 5x^2
Again, to determine the x-intercepts, we set y equal to zero:
0 = 2x^4 - 5x^2
To solve for x, we can factor out the common term x^2:
0 = x^2(2x^2 - 5)
Now, we have two possible solutions:
x^2 = 0 => x = 0 (double root - it intersects the x-axis at x = 0)
2x^2 - 5 = 0
To solve the quadratic equation 2x^2 - 5 = 0, we can apply the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 2, b = 0, and c = -5. Plugging in the values:
x = (-0 ± √(0^2 - 4*2*(-5))) / (2*2)
Simplifying:
x = (± √(0 + 40)) / 4
x = (± √40) / 4
x = (± √(4 * 10)) / 4
x = (± 2√10) / 4
x = ± √10 / 2
This gives us two solutions:
x = √10 / 2
x = -√10 / 2
Therefore, the x-intercepts for the equation y = 2x^4 - 5x^2 are x = 0, x = √10 / 2, and x = -√10 / 2.