To find the function \( g(x) \) that represents the translation of \( f(x) = x^2 \) by 5 units left and 1 unit down, we start by understanding how to apply translations to a function.
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Translation 5 units left: To move a function left by \( h \) units, we replace \( x \) with \( x + h \). For a translation of 5 units left, we replace \( x \) with \( x + 5 \): \[ f(x) = x^2 \quad \Rightarrow \quad f(x + 5) = (x + 5)^2 \]
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Translation 1 unit down: To move a function down by \( k \) units, we subtract \( k \) from the function's output. Here, \( k = 1 \): \[ g(x) = f(x + 5) - 1 = (x + 5)^2 - 1 \]
Now, we can simplify \( (x + 5)^2 - 1 \): \[ (x + 5)^2 = x^2 + 10x + 25 \] Thus, \[ g(x) = x^2 + 10x + 25 - 1 = x^2 + 10x + 24 \]
Next, we want to express \( g(x) \) in the vertex form \( a(x-h)^2 + k \). We can complete the square for the quadratic \( x^2 + 10x + 24 \):
- Take the coefficient of \( x \), which is \( 10 \), divide it by \( 2 \) to get \( 5 \), and square it to obtain \( 25 \).
- Rewrite the expression: \[ g(x) = x^2 + 10x + 25 - 1 = (x + 5)^2 - 1 \]
So, we have: \[ g(x) = 1(x + 5)^2 - 1 \]
In the form \( a(x-h)^2 + k \), we can identify:
- \( a = 1 \)
- \( h = -5 \) (so, \( h \) is actually \( -5 \))
- \( k = -1 \)
Thus, the final form of \( g(x) \) is: \[ g(x) = 1(x + 5)^2 - 1 \]
Therefore: \[ \boxed{g(x) = (x + 5)^2 - 1} \]