Question

Y=-3×2+8×+35.Identify the axis of symmetry and the vertex?

Answer 1

`"Vertex: " (4/3, 363/9)`
`"Axis of Symmetry: " x=4/3`

Explanation:

`y=-3x^2+8x+35`

It's important to remember that, when it comes to quadratics, there are two forms:

`f(x)=ax^2+bx+c` `color(blue)(" Standard Form")`
`f(x)=a(x-h)^2+k` `color(blue)(" Vertex Form")`

For this problem, we can disregard the vertex form, as our equation is in the standard form.

To find the vertex of the standard form, we have to do some math:

`"Vertex:"` `((-b)/(2a), f((-b)/(2a)))`

The `y"-coordinate"` might look a little confusing, but all it means is that you plug in the `x"-coordinate"` of the vertex back into the equation and solve. You'll see what I mean:

`x"-coordinate:"`

`((-b)/(2a))`

`((-8)/(2(-3)))` `color(blue)(" Plug in "8 " for " b " and " -3 " for " a)`

`((-8)/-6)` `color(blue)(" "2*3=6)`

`((cancel(-)4)/(cancel(-)3))` `color(blue)(" Simplify; negatives cancel to make positive")`

`x"-coordinate: " color(red)(4/3)`

Now let's plug `4/3` back into every `x` in the original function

`y=-3x^2+8x+35`

`y=-3(4/3)^2+8(4/3)+35` `color(blue)(" Plug " 4/3 " into the " x"'s")`

`y= -3(16/9)+8(4/3)+35` `color(blue)(" "4^2=16," " 3^2=9)`

`y=-48/9 +8(4/3)+35` `color(blue)(" "-3*16=-48)`

`y=-48/9+32/3+35` `color(blue)(" "8*4=32)`

Let's get some common denominators to simplify this:

`y=-48/9+96/9+35` `color(blue)(" "32*3=96," "3*3=9)`

`y=-48/9+96/9+315/9` `color(blue)(" "35*9=315," "1*9=9)`

`y=48/9+315/9` `color(blue)(" "-48/9+96/9=48/9)`

`y=363/9` `color(blue)(" "48/9+315/9=363/9)`

`y"-coordinate: " color(red)(363/9)`

Now that we have our `x` and `y` `"coordinates,"` we know the vertex:

`"Vertex: " color(red)((4/3, 363/9)`

When it comes to quadratics, the `"axis of symmetry"` is always the `x"-coordinate"` of the `"vertex"`. Therefore:

`"Axis of Symmetry: " color(red)(x=4/3)`

It's important to remember that the `"axis of symmetry"` is always told in terms of `x`.

Answer 2

`x=4/3," vertex "=(4/3,121/3)`

Explanation:

`"the equation of a parabola in "color(blue)"vertex form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(y=a(x-h)^2+k)color(white)(2/2)|)))`

`"where "(h,k)" are the coordinates of the vertex and a"`
`"is a multiplier"`

`"to express y in this form use "color(blue)"completing the square"`

`• " the coefficient of the "x^2" term must be 1"`

`rArry=-3(x^2-8/3x-35/3)`

`• " add/subtract "(1/2"coefficient of the x-term")^2" to"`
`x^2-8/3x`

`y=-3(x^2+2(-4/3)xcolor(red)(+16/9)color(red)(-16/9)-35/3)`

`color(white)(y)=-3(x-4/3)^2-3(-16/9-35/3)`

`color(white)(y)=-3(x-4/3)^2+121/3larrcolor(red)"in vertex form"`

`rArrcolor(magenta)"vertex "=(4/3,121/3)`

`"the equation of the axis of symmetry passes through the"`
`"vertex is vertical with equation "x=4/3`