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zavuch27 [327]
3 years ago
10

Select all the statements about functions g(x) = x2 − 4x + 3 and f(x) = x2 − 4x that are true. A. The vertex of the graph of fun

ction g is above the vertex of the graph of function f. B. The graphs have the same axis of symmetry. C. Function f has a maximum value and function g has a minimum value. D. The graphs never intersect. E. The graphs have the same y-intercept.

Mathematics
2 answers:
Verizon [17]3 years ago
6 0

9514 1404 393

Answer:

  A, B, D

Step-by-step explanation:

We observe that g(x) = f(x) +3, which means it is the graph of f(x) translated up 3 units. That translation does not change the axis of symmetry, but it puts the vertex of g above the vertex of f (by 3 units).

Since each point on the graph of g is 3 units above the corresponding point of f, the graphs cannot intersect.

The true statements are ...

A. The vertex of the graph of function g is above the vertex of the graph of function f.

B. The graphs have the same axis of symmetry.

D. The graphs never intersect.

QveST [7]3 years ago
4 0

Answer:

,

Step-by-step explanation:

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musickatia [10]

<u>ANSWER:</u>

Square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

<u>SOLUTION: </u>

Given, we have to find square roots of 5 + 12i and 5 – 12i

<u><em>a) 5 + 12i</em></u>

Now, square root of 5 + 12i  

Suppose that a + bi is a square root of 5 + 12i.  

\text { Then, }(a+b i)^{2}=\left(a^{2}-b^{2}\right)+(2 a b) i=5+12 i

Equate real and imaginary parts:

\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {\text { So, } a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0}\end{array}

Since a must be real, a = 3 or -3.  

This gives b = 2 or -2, respectively.  

Thus, we have two square roots: 3+2i or -3-2i.

<u><em>b) 5 - 12i</em></u>

Now, square root of 5 - 12i  

Suppose that a - bi is a square root of 5 - 12i.  

\text { Then, }(a-b i)^{2}=\left(a^{2}-b^{2}\right)-(2 a b) i=5-12 i

Equate real and imaginary parts:

\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {S o, a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0} \\\\ {\text { since a must be real, } a=3 \text { or }-3}\end{array}

This gives b = 2 or -2, respectively.  

Thus, we have two square roots: 3 - 2i or -3 + 2i.

Hence, square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

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