concerning the graphs of a function

**loes** · March 24th 2009, 01:42 PM

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?

**Grandad** · March 25th 2009, 09:19 AM

Hello loes

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

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

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

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

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

Where is the vertex of $y = x^2$ ? So where will it end up when you apply all three of these transformations?

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

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

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

what is the image set of the function in interval notation?

What is the image set for $y = x^2$ ? What will happen to this set when the transformations are applied?

Grandad

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