# concerning the graphs of a function

• Mar 24th 2009, 01:42 PM
loes
concerning the graphs of a function
f(x)= 1/4(x-2)^2-4

how can the parabola that is the graph of f be obtained from the graph of y=x^2

what are the co-ordinates of the vertex of the parabola?

what are the x-intercepts and the y-intercepts of the parabola?

what is the image set of the function in interval notation?
• Mar 25th 2009, 09:19 AM
Graphical interpretation of Transformation of a function
Hello loes
Quote:

Originally Posted by loes
f(x)= 1/4(x-2)^2-4

how can the parabola that is the graph of f be obtained from the graph of y=x^2

The graph of \$\displaystyle f(x - a)\$ is the graph of \$\displaystyle f(x)\$ shifted to the right \$\displaystyle a\$ units.

The graph of \$\displaystyle kf(x)\$ is the graph of \$\displaystyle f(x)\$ stretched parallel to the \$\displaystyle y\$-axis with factor \$\displaystyle k\$.

The graph of \$\displaystyle f(x) - b\$ is the graph of \$\displaystyle f(x)\$ shifted downwards \$\displaystyle b\$ units.

Put these three facts together, and you can describe how the graph of \$\displaystyle y=\tfrac{1}{4}(x-2)^2 - 4\$ is obtained from the graph of \$\displaystyle y = x^2\$.

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what are the co-ordinates of the vertex of the parabola?
Where is the vertex of \$\displaystyle y = x^2\$? So where will it end up when you apply all three of these transformations?

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what are the x-intercepts and the y-intercepts of the parabola?
The \$\displaystyle x\$-intercepts are the values of \$\displaystyle x\$ that satisfy \$\displaystyle \tfrac{1}{4}(x-2)^2 - 4=0\$.

The \$\displaystyle y\$-intercept is the value of \$\displaystyle y\$ when \$\displaystyle x = 0\$.

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what is the image set of the function in interval notation?
What is the image set for \$\displaystyle y = x^2\$? What will happen to this set when the transformations are applied?