Yes, (3, 2) is a solution to y = 2x - 4. (:
Answer:
B
Step-by-step explanation:
Using the determinant to determine the type of zeros
Given
f(x) = ax² + bx + c ( a ≠ 0 ) ← in standard form, then the discriminant is
Δ = b² - 4ac
• If b² - 4ac > 0 then 2 real and distinct zeros
• If b² - 4ac = 0 then 2 real and equal zeros
• If b² - 4ac < 0 then 2 complex zeros
Given
f(x) = (x - 1)² + 1 ← expand factor and simplify
= x² - 2x + 1 + 1
= x² - 2x + 2 ← in standard form
with a = 1, b = - 2, c = 2, then
b² - 4ac = (- 2)² - (4 × 1 × 2) = 4 - 8 = - 4
Since b² - 4ac < 0 then the zeros are complex
Thus P(x) has no real zeros
Answer:
-2+4i
Step-by-step explanation:
Simplify by combining the real and imaginary parts of each expression.
You can use lattice multiplication or long multiplication. for examples and/or visuals message me.
Answer:
The end behavior of a polynomial is determined by its grade, if it's an odd polynomial or an even polynomial. Also, another important characterstic is the leading coeffcient, because if it's positive, then the function will open to positive territory, this apply the most to even polynomials.
The given expression is
As you can notice, the polynomial function has a grade of 4, that means it's an even function, with negative leading coefficient.
Therefore, the given function open towards negative side, that means it's end behavior is downside.
The image attached shows the graph of the function, there you can observe the end behavior we determined.
