Question

How do you find the important points to graph `y = (x - 3)² + 1`?

Answer 1

See explanation

Explanation:

Standard form of the equation is
`y=ax^2+bx+c`
Note that `a` could be of value 1
If `a` is positive then the graph is a horse shoe shape with the curve at the bottom. If negative then the other way round.

Square the bracket and then add the 1 giving:
`y=x^2 - 6x +4`

Using standard form equation

`x" of "(x,y)_("minimum")" is "(-1/2) times b/a`

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So for your case we have:

`color(green)(x" of "(x,y)_("minimum") =(-1/2) times (-6) = color(green)(+3))`

`color(red)("substitute "x=3" in your equation to find "y_("minimum"))`

`y = (3)^2 -6(3)+4`
`y =13 - 18`
`y = -5`
`color(green)(y" of " (x,y)_("minimum")=(-5)`

Putting it all together

`color(green)( (x,y)_("minimum")=(3,-5)`
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`color(green)("To find x-intercepts substitute y=0 and solve")`

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`color(green)("To find y-intercepts substitute x=0 and solve")`

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