Question

Question #17e3c

Answer 1

`color(magenta)("Solution part 1 of 2")`

Full explanation given so a little long

Explanation:

You also need the x-intercepts

`color(blue)("The general shape of the graph")`

As `-2x^2` is negative the graph is of shape type `nn`

IF it had been `+2x^2`, ie positive, then the graph would have been of type `uu`
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`color(blue)("Determine "x_("vertex") ->" axis of symmetry")`

`color(brown)("Using part of the process for completing the square.")`

See my example https://socratic.org/s/aEmV2Z2e for the complete process (different values)

General case: given that `y=ax^2+bx+c`

we need `" "y=a(x^2+b/ax)+c`

`x_("vertex")=(-1/2)xxb/a`

So for this question: `y=color(red)(-2)(x^2+color(green)(4/(2))x)+1`

Note that `color(red)((-2))xx color(green)(4/2)x = -4x` as required

`" "color(blue)(ul(bar(|" "x_("vertex")=(-1/2)xx(4/2)=-1" "|))`

`" "color(blue)(ul(bar(|color(white)(2/2)"Axis of symmetry "->x=-1" "|))`
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`color(blue)("Determine "y_("vertex"))`

Substitute for `x=-1`

`y=-2x^2-4x+1" "->" "y=-2(-1)^2-4(-1)+1`

`" "->" "y=" "-2" "+4" "+1`

`" "color(blue)(ul(bar(|color(white)(2/2)y_("vertex")=+3" "|))`
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`" "color(blue)(ul(bar(|color(white)(2/2)"Vertex "->(x,y)=(-1,+3)" "|))`
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`color(blue)("Determine y-intercept")`

Consider the given equation: `y=-2x^2-4xcolor(red)(+1)`

`" "color(blue)(ul(bar(|color(white)(2/2)y_("intercept")=color(red)(+1)" "|))`

Why is this? The graph crosses the y-axis at `x=0`

`y=-2x^2-4x+1" "->" "y=-2(0)^2-4(0)+1`
`" "->" "y= " "0" "-0color(white)(.)+1`
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`color(blue)("Determine the x-intercepts")`

Running out of space on this solution so adding another one as a continuation:

`color(magenta)("See solution 2 of 2")`

Answer 2

`color(magenta)("Solution part 2 of 2")`

Determine the x-intercepts

Again; full explanation given

Explanation:

You can either use the 'complete the square' method or use the standard formula approach

Complete the square `->` https://socratic.org/s/aEmWR5k7
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As the graph is of shape `nn` and the vertex is at `y=3` the graph crosses the x-axis so x-intercepts exist.

Given that `y-ax^2+bx+c` then

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

`y=-2x^2-4x+1` gives:

`a=-2"; "b=-4"; "c=+1`

Substitute and solve for `x`

OR YOU CAN FACTORISE (not always easy)

`y=(-2x+-?)(x+-?) = -2x^2+.....` That part works!

`y=(-2x+1)(x+1) = -2x^2-2x+x+1` This is just not going to work so we are back to the formula.

`x=(+4+-sqrt((-4)^2-4(-2)(+1)))/(2(-2))`

`x=4/(-4)+-sqrt(16+8)/(-4)`

`x=-1+-sqrt(24)/(-4)`

But 24 = `2xx3xx2^2` so `sqrt(24)=2sqrt(6)`

`x=-1+-(2sqrt(6))/(-4)`

`x=-1+-sqrt(6)/2` Exact values

`x~~-2.225`
`x~~+0.225`

All to 2 decimal places

Tony B