204k views
2 votes
F(x) = x2 – 3x – 2 is shifted 4 units left. The result is g(x). What is g(x)?

User Rassar
by
7.9k points

2 Answers

0 votes

\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}


\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}


\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}

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


\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)


User Darius
by
7.9k points
2 votes

Answer:

The required function is
g(x)=x^2+5x+2.

Explanation:

The given function is


f(x)=x^2-3x-2

The transformation of a function is defied as


g(x)=f(x+a)+b

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.


g(x)=f(x+4)


g(x)=(x+4)^2-3(x+4)-2


g(x)=x^2+8x+16-3x-12-2


g(x)=x^2+5x+2

Therefore the required function is
g(x)=x^2+5x+2.

User Abhik
by
8.5k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories