Question

How do you find the y intercept, axis of symmetry and the vertex to graph the function `f(x)=2x^2+5x`?

Answer 1

`y_"intercept"->(x,y)=(0,0)`

`x_"intercepts"=0 and x=-5/2`

Vertex`->(x,y)=(-5/4,-25/8)`

Explanation:

`color(blue)("Determine the y-intercept")`

Write as: `y=2x^2+5x+0`

The y-intercept is the value of the constant so this is at `y=0`

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`color(blue)("Determine the vertex by completing the square")`

given the standard form of `y=ax^2+bx+c`

Start by writing as:

`y=a(x^2+b/a x)+c+k" "->" "y=2(x^2+5/2x)+0+k`

Where `k` is a correction for an error this process introduces
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Move the square root to outside the brackets:

`y=a(x+b/a x)^2+c+k" "->" "y=2(x+5/2x)^2+0+k`
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Discard the `x` from `b/ax`

`y=a(x+b/a )^2+c+k" "->" "y=2(x+5/2)^2+0+k`
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Halve the `b/a`

`y=a(x+b/(2a) )^2+c+k" "->" "y=2(x+5/4)^2+0+k`
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The error comes from: `a(???+b/(2a))^2 = a(b^2/(4a^2))=b^2/(4a)`

Set `b^2/(4a)+k=0" "->" "2(25/16)+k=0`

`=>k=-25/8`
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`y=a(x+b/(2a) )^2+c+k" "->" "y=2(x+5/4)^2-25/8`

`x_"vertex"=(-1)xxb/(2a)" "->" "(-1)xx(+5/4)=-5/4`
`y_"vertex"=c+k" "->" 0-25/8`

Vertex`->(x,y)=(-5/4,-25/8)`
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`color(blue)("Determine x-intercepts")`

Using: `y=2(x+5/4)^2-25/8`

Set `y=0`

Add `25/8` to both sides

`+25/8=2(x+5/4)^2`

Divide both sides by 2

`25/16=(x+5/4)^2`

Square root both sides

`+-sqrt(25/16)-5/4=x`

`x=5/4+-5/4`

`x=0 and x=-5/2`

Tony B