Respuesta :
Answer:
A. [tex](2x-3)(x-1)[/tex]
B. [tex](1,0) \\(1.5,0)[/tex]
[tex]2x^2-5x+3\\\frac{5(+ or -)\sqrt{(-5)^2-(4)(2)(3)} }{2(2)} \\\frac{5(+ or -)\sqrt{25-24} }{4}\\\frac{5(+ or -)\sqrt{1} }{4}\\ \frac{5(+ or -)1 }{4}\\\frac{5+1}{4} or \frac{5-1}{4} \\ \frac{6}{4} =1.5 and \frac{4}{4} =1[/tex]
C. Both sides will continue to increase as they go because this is an even degree quadratic equation.
D. You use the two x-intercepts to make sure it is in the correct area and since we know the end behavior all you'd have to assume is that it continues to go up so any negative answers you may accidentally receive would be incorrect.
Part A: The completely factorized function is f(x) = (2x - 3)(x - 1).
Part B: The x-intercepts of the graph f(x) are (1, 0) and (1.5, 0).
Part C: The end behavior is that f(x) → ∞ as x → -∞, f(x) → ∞ as x → ∞.
Part D: The graph of f(x) is attached.
How do we solve the given question?
The given function is f(x) = 2x² - 5x + 3
Part A: Factor f(x) completely.
To factor f(x), we use the mid-term factorization method.
f(x) = 2x² - 5x + 3
or, f(x) = 2x² - 2x - 3x + 3
or, f(x) = 2x(x-1) -3(x-1)
or, f(x) = (2x-3)(x-1)
∴ The completely factorized function is f(x) = (2x - 3)(x - 1).
Part B: To find x-intercepts.
To find x-intercepts, we take f(x) = 0.
∴ (2x - 3)(x - 1) = 0
So the x-intercepts are given by,
2x - 3 = 0 ⇒ x = 3/2 = 1.5
x - 1 = 0 ⇒ x = 1
∴ The x-intercepts of the graph f(x) are (1, 0) and (1.5, 0).
Part C: The end behavior of the graph f(x).
Since the leading term of the polynomial (the term in the polynomial which contains the highest power of the variable) is 2x², the degree is 2, that is, even, and the leading coefficient is 2, that is, positive.
∴ The end behavior is that f(x) → ∞ as x → -∞, f(x) → ∞ as x → ∞.
Part D: Steps to draw the graph f(x).
We draw the graphs using the conclusions from Part B and Part C.
We plot the points from part B: (1, 0) and (1.5, 0) and then make the end behavior of the graph the same as what we got in part C.
The graph is attached.
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