What does it mean for an equation to have no solution or infinite solutions? How does the equation appear when those are the out
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Short Answer f(x) down 5 units, left 2 units and right side up from g(x)
Step One
Find out what -g(x+2) is.
There are two steps to this. The first is to deal with the minus sign
(g(x)) = - x^2 + 5. Be very careful what you do next.
Put brackets around both terms on the right.
g(x) = (-x^2 + 5) Now put the minus sign in front of g(x) and another one in front of the brackets.
-g(x) = - (-x^2 + 5) Remove the brackets.
-g(x) = x^2 - 5
<em>Result</em>
So far g(x) moves down 5 units and opens upward.
Step Two
The second step is to see what the x + 2 does.
-g(x + 2) = (x + 2)^2 - 5
The final result is that the graph opens upward, moves left 2 spaces and down 5.
There is a graph enclosed to show you the key steps.
The red graph is g(x) = x^2 - 5
The blue graph is g(x+2) = (x+2)^2 - 5
Step One
Find out what -g(x+2) is.
There are two steps to this. The first is to deal with the minus sign
(g(x)) = - x^2 + 5. Be very careful what you do next.
Put brackets around both terms on the right.
g(x) = (-x^2 + 5) Now put the minus sign in front of g(x) and another one in front of the brackets.
-g(x) = - (-x^2 + 5) Remove the brackets.
-g(x) = x^2 - 5
<em>Result</em>
So far g(x) moves down 5 units and opens upward.
Step Two
The second step is to see what the x + 2 does.
-g(x + 2) = (x + 2)^2 - 5
The final result is that the graph opens upward, moves left 2 spaces and down 5.
There is a graph enclosed to show you the key steps.
The red graph is g(x) = x^2 - 5
The blue graph is g(x+2) = (x+2)^2 - 5