How is the graph of $y = \frac{1}{3} x^{2} - 4$ $y=1/3x^2-4$ related to the graph of $f (x) = x^{2}$ $f(x)=x^2$ ?

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How is the graph of $y = \frac{1}{3} x^{2} - 4$ $y=1/3x^2-4$ related to the graph of $f (x) = x^{2}$ $f(x)=x^2$ ?

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Answer 1 · 2017-12-23T15:05:02

The lines intersect
$(\sqrt{6}, 6) (- \sqrt{6}, 6)$ $(sqrt(6),6)(-sqrt(6),6)$

Explanation:

f(x) is the expression used for functions. It is another way of showing the y value because it depends on the variable x. The y value is created after the x value is put through the function

Thus

$f (x) = x^{2}$ $f(x) = x^2$ is the same as $y = x^{2}$ $y = x^2$

Both equation have y as their subjects. This way an inequality can be created.

$x^{2} = \frac{1}{3} x^{2} - 4$ $x^2=1/3x^2-4$

Make x the subject

$x^{2} + 4 = \frac{1}{3} x^{2}$ $x^2+4=1/3x^2$

$3 x^{2} + 12 = x^{2}$ $3x^2+12=x^2$

$2 x^{2} = - 12$ $2x^2 = -12$

$x^{2} = - 6$ $x^2=-6$

$x = \pm \sqrt{6}$ $x=+-sqrt(6)$

Once the x coordinate has been found, the y one can be found two. There are two values of x so there are also two coordinates.

$(\sqrt{6}, 6) (- \sqrt{6}, 6)$ $(sqrt(6),6)(-sqrt(6),6)$

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How is the graph of $y = \frac{1}{3} x^{2} - 4$ $y=1/3x^2-4$ related to the graph of $f (x) = x^{2}$ $f(x)=x^2$ ?

1 Answer

Explanation:

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How is the graph of $y = \frac{1}{3} x^{2} - 4$ $y=1/3x^2-4$ related to the graph of $f (x) = x^{2}$ $f(x)=x^2$ ?