9514 1404 393
Answer:
(a) none of the above
Step-by-step explanation:
The largest exponent in the function shown is 2. That makes it a 2nd-degree function, also called a quadratic function. The graph of such a function is a parabola -- a U-shaped curve.
The coefficient of the highest-degree term is the "leading coefficient." In this case, that is the coefficient of the x² term, which is 1. When the leading coefficient of an even-degree function is positive, the U curve has its open end at the top of the graph. We say it "opens upward." (When the leading coefficient is negative, the curve opens downward.)
This means the bottom of the U is the minimum value the function has. For a quadratic in the form ax²+bx+c, the horizontal location of the minimum on the graph is at x=-b/(2a). This extreme point on the curve is called the "vertex."
This function has a=1, b=1, and c=3. The minimum of the function is where ...
x = -b/(2·a) = -1/(2·1) = -1/2
This value is not listed among the answer choices, so the correct choice for this function is ...
none of the above
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The attached graph of the function confirms that the minimum is located at x=-1/2
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<em>Additional comment</em>
When you're studying quadratic functions, there are few formulas that you might want to keep handy. The formula for the location of the vertex is one of them.
Answer:
A.( 3m+28)/4
Step-by-step explanation:
So basically since it is addition we just add 9 to -2 to get 7. So then the equation becomes (3m/4)+7. Then we will multiply 7 by 4 to get common denominator so it will be (3m+(7*4))/4 which is (3m+28)/4.
So the division of a number and -9 or
x/-9
+10=11 or
(x/-9)+10=11
subtract 10 from both sides
x/-9=1
multiply both sides by -9
x=-9
A. True. We see this by taking the highest order term in each factor:

B. True. Again we look at the leading term's degree and coefficient. f(x) behaves like -3x⁶ when x gets large. The degree is even, so as x goes to either ± ∞, x⁶ will make it positive, but multiplying by -3 will make it negative. So on both sides f(x) approaches -∞.
C. False. f(x) = 0 only for x=0, x = 5, and x = -2.
D. False. Part of this we know from the end behavior discussed in part B. On any closed interval, every polynomial is bounded, so that for any x in [-2, 5], f(x) cannot attain every positive real number.
E. True. x = 0 is a root, so f(0) = 0 and the graph of f(x) passes through (0, 0).
F. False. (0, 2) corresponds to x = 0 and f(x) = 2. But f(0) = 0 ≠ 2.