Respuesta :
[tex]\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}[/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative} \\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \end{array}[/tex]
[tex]\bf \begin{array}{llll} \bullet \textit{ vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}} \end{array}[/tex]
now... your expression is not in vertex form, that's ok... to do a horizontal shift to the left by 4 units, we can simply, add the C and B component to the "x" variable, C=4, B =1, that way the horizontal shift of C/B or 4/1 is just +4, giving us a horizontal shift to the left
[tex]\bf f(x)=x^2-3x-2\impliedby \textit{let's change that for }f(1x+4) \\\\\\ f(1x+4)=(1x+4)^2-3(1x+4)-2 \\\\\\ f(x+4)=x^2+8x+16-3x-12-2 \\\\\\ f(x+4)=x^2-5x+2\impliedby g(x)[/tex]
[tex]\bf \begin{array}{llll} % right side info \bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\ \bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative} \\\\ \bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\ \qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\ \end{array}[/tex]
[tex]\bf \begin{array}{llll} \bullet \textit{ vertical shift by }{{ D}}\\ \qquad if\ {{ D}}\textit{ is negative, downwards}\\\\ \qquad if\ {{ D}}\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{{{ B}}} \end{array}[/tex]
now... your expression is not in vertex form, that's ok... to do a horizontal shift to the left by 4 units, we can simply, add the C and B component to the "x" variable, C=4, B =1, that way the horizontal shift of C/B or 4/1 is just +4, giving us a horizontal shift to the left
[tex]\bf f(x)=x^2-3x-2\impliedby \textit{let's change that for }f(1x+4) \\\\\\ f(1x+4)=(1x+4)^2-3(1x+4)-2 \\\\\\ f(x+4)=x^2+8x+16-3x-12-2 \\\\\\ f(x+4)=x^2-5x+2\impliedby g(x)[/tex]
Answer:
The required function is [tex]g(x)=x^2+5x+2[/tex].
Step-by-step explanation:
The given function is
[tex]f(x)=x^2-3x-2[/tex]
The transformation of a function is defied as
[tex]g(x)=f(x+a)+b[/tex]
Where a is horizontal shift and b is the vertical shift.
If a>0, then the graph of f(x) shifts a units left and if a<0, then the graph of f(x) shifts a units right.
Since the graph of f(x) shifts 4 units left, therefore a=4.
[tex]g(x)=f(x+4)[/tex]
[tex]g(x)=(x+4)^2-3(x+4)-2[/tex]
[tex]g(x)=x^2+8x+16-3x-12-2[/tex]
[tex]g(x)=x^2+5x+2[/tex]
Therefore the required function is [tex]g(x)=x^2+5x+2[/tex].
