The vertex of the function f(x) exists (1, 5), the vertex of the function g(x) exists (-2, -3), and the vertex of the function f(x) exists maximum and the vertex of the function g(x) exists minimum.
<h3>How to determine the vertex for each function is a minimum or a maximum? </h3>
Given:
and
The generalized equation of a parabola in the vertex form exists
Vertex of the function f(x) exists (1, 5).
Vertex of the function g(x) exists (-2, -3).
Now, if (a > 0) then the vertex of the function exists minimum, and if (a < 0) then the vertex of the function exists maximum.
The vertex of the function f(x) exists at a maximum and the vertex of the function g(x) exists at a minimum.
To learn more about the vertex of the function refer to:
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I just did the second one, the one that looks this a side ways L,
here are the points
(4,4). (5,4). (3,1). (3,2). (4,2). hope this helped
17) f(x) = 16/(13-x).
In order to find domain, we need to set denominator expression equal to 0 and solve for x.
And that would be excluded value of domain.
13-x =0
Adding x on both sides, we get
13-x +x = x.
13=x.
Therefore, domain is All real numbers except 13.
18).f(x) = (x-4)(x+9)/(x^2-1).
In order to find the vertical asymptote, set denominator equal to 0 and solve for x.
x^2 -1 = 0
x^2 -1^2 = 0.
Factoring out
(x-1)(x+1) =0.
x-1=0 and x+1 =0.
x=1 and x=-1.
Therefore, Vertical asymptote would be
x=1 and x=-1
19) f(x) = (7x^2-3x-9)/(2x^2-4x+5)
We have degrees of numberator and denominator are same.
Therefore, Horizontal asymptote is the fraction of leading coefficents.
That is 7/2.
20) f(x)=(x^2+3x-2)/(x-2).
The degree of numerator is 2 and degree of denominator is 1.
2>1.
Degree of numerator > degree of denominator .
Therefore, there would no any Horizontal asymptote.
Intersection
Where the graph line matches with which column of graph...... take a line from that place to the axis to identify value on that point of intersection
Hope it helped !!
