The graph of the function f(x) =ax2+bc+c is given below. Could the factors of f(x) be (x+2) and (x+4)? Based on the graph, expla
in why or why not.
1 answer:
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Answer:
Step-by-step explanation:
Slope of line passing through and
is given by,
m =
From the graph attached,
Slope of the line passing through (-3, 0) and (0, 5) =
=
Line passing through a point (h, k) and slope 'm' will be,
y - h = m(x - k)
Since, line passing through (11, 0) is parallel to the line given in the graph,
Slope of the parallel line will be same as
Equation of a line passing through (11, 0) and slope =
y - 0 =
y =
Now satisfy this equation with the points given in the options,
Option (1)
For (1.67, -15.59),
-15.59 =
-15.59 = -15.55
False.
Therefore, given point doesn't lie on the line.
Option (2)
For (-0.33, -13.59)
-13.59 =
-13.59 = -18.88
False
Option (3)
For (0.67, -18.59),
-18.59 =
-18.59 = 17.22
False
Option (4)
For (1.67, -15.59)
-15.59 =
-15.59 = -15.55
False.
Therefore, none of the points given in the options are lying on the line.
First lets start with number six. The only way to solve this is if you determine what "a" and "b" are using the first log they have given to you.
The first variable that I solved for was "a" and }
The same is also true for "b", but when you put both "a" and "b" together the only combination that I have found to work is
Next you plug these numbers in for "a" and "b" on the second equation to get something that looks like this: and the picture below shows where the answer becomes a negative fraction.
https://www.symbolab.com/solver/logarithms-calculator/%20%5Clog_%7B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%5Cfrac%7B1%7D%7B4%7D%7D%5Cleft(%5Cfrac%7B%5Csqrt%7B%5Cfrac%7B1%7D%7B2%7D%7D%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B4%7D%7D%7D%5Cright)
If you paste that link in your search bar it will give you a even more in depth understanding of how to get this answer
Next is #7, the easier of the two.There are two ways to solve for your answers. According to the graph of this equation there are four possible real solutions. . (This does not account for any complex solutions)
Notice that the bases are conjugates which is why the answers are so "nice"
The key is in the exponents
if then the sum on the conjugates will be 10 so
so
Now for the other two
the solution is also true if
so
the four real solutions are
