Question

How is the graph of `g(x)=-5-4/3x^2` related to the graph of `f(x)=x^2`?

Answer 1

See explanation

Explanation:

Set `y_1=g(x)=-4/3x^2+0x-5`

Set `y_2=f(x)=x^2+0x+0`

For `f(x)`
If the `x^2` term is positive then the graph is of general shape `uuu`

For `g(x)`
If the `x^2` term is negative then the graph is of general shape `nnn`

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The `4/3` bit from `-4/3x` makes the graph 'narrower' as `4/3>1`
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The graph passes through the y-axis at `y=c` that is:
`(x,y)=(0,-5)`
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For the vertex of the graph we do this (part of completing the square):

Write as `color(green)(y_1=-4/3(x^2color(red)(-(3/4)0)x)-5)`

The `x_("vertex")=(-1/2)xx[color(red)(-(3/4)0)]`

`x_("vertex")=-1/2xx0 =0`

As in both `f(x) and g(x)` we have `0x` so the axis of symmetry is the y-axis (`x_("vertex")=0`)

`f(x)` axis of symmetry is the y-axis. The constant is 0 so the vertex is at `(x,y)->(0,0)`

`g(x)` axis of symmetry is the y-axis. The constant is -5 so the vertex is at `(x,y)->(0,-5)`

Answer 2

`f(x)=x^2` is the parent function of `g(x)=-5-4/3x^2`

Explanation:

First, convert to vertex form:

`g(x)=-4/3x^2-5`

It is already in vertex form, or you could put it as

`g(x)=-4/3(x-0)^2-5`

Now, we can tell the parent function of this equation is `f(x)=x^2`

The specific transformations are:

`"Reflect over x-axis"`
`"Vertical Stretch by 4/3"`
`"Down 5"`