Question

How do you find the vertex and the intercepts for `4x^2 - 3x - 2 = 0`?

Answer 1

`color(blue)("at "y=0 ....x=(3+sqrt(23))/4 and x=(3-sqrt(23))/4)`

`color(blue)("vertex"->(x,y)->(3/8,-41/16))`

`color(blue)(y_("intercept")=-2`

Explanation:

There is a really cool way to find the axis of symmetry and hence `x_("vertex")`. Some you can do in your head!

Write the given equation as:`" "y=4(x^2-3/4x)-2`

Now consider the `-3/4` from `-3/4x`

Apply:`" " (-1/2)xx(-3/4)=+ 3/8`

`color(blue)(x_("vertex")=3/8)`
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`color(blue)("Determin "y_("vertex"))`
Substitute `x=3/8` in the original equation

`y_("vertex")= 4(3/8)^2-3(3/8)-2`

`y_("vertex")= 9/16-9/8-2 = -2 9/16 ->-41/16`

`color(blue)(y_("vertex")= -2 9/16 ->-41/16)`

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`color(blue)("Determine "y_("intercept")`

The graph will cross the y-axis at x=0 so by substitution.

`y=4(0)^2-3(0)-2`

`y=0-0-2`

`color(blue)(y_("intercept")=-2`
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`color(blue)("Determine "x_("intercept"))`

There are not whole number factors for this equation so we are forced to use the equation method.

`y=ax^2+bx+c`

Where`" "x=(-b+-sqrt(b^2-4ac))/(2a)`

and:`a=4; b=-3; c=-2 " "`giving

`x=(+3+-sqrt((-3)^2-4(4)(-2)))/(2(4))`

`x=(3+-sqrt(23))/4`

23 is a prime number so you can not do any more with it. So if you whish have a precise solution there is no choice other than to state:

`color(blue)("at "y=0 ....x=(3+sqrt(23))/4 and x=(3-sqrt(23))/4)`

You could work out the decimals if you wish but these will not be precise!

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