Question

How do you find the vertex and intercepts for `y = (1/8)(x – 5)^2 - 3`?

Answer 1

`"see explanation"`

Explanation:

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

`•color(white)(x)y=a(x-h)^2+k`

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

`y=1/8(x-5)^2-3" is in this form"`

`color(magenta)"vertex "=(5,-3)`

`"to find y-intercept let x = 0"`

`y=25/8-24/8=1/8larrcolor(red)"y-intercept"`

`"to find x-intercepts let y = 0"`

`1/8(x-5)^2-3=0`

`1/8(x-5)^2=3rArr(x-5)^2=24`

`color(blue)"take the square root of both sides"`

`x-5=+-sqrt24=+-2sqrt6`

`"add 5 to both sides"`

`x=5+-2sqrt6larrcolor(red)"exact solutions"`

`x~~0.1,x~~9.9" to 1 dec. place"`

Answer 2

Vertex : `(5, -3)`,

x-intercepts : `=5\pm2\sqrt6` &

y-intercept: `=1/8`

Explanation:

Given equation:

`y=(1/8)(x-5)^2-3`

`1/8(x-5)^2=y+3`

`(x-5)^2=8(y+3)`

The above equation is in form of vertical parabola: `(x-x_1)^2=4a(y-y_1)` which has

Vertex `(x_1, y_1)\equiv(5, -3)`

Setting `x=0` in the above equation of parabola:

`(0-5)^2=8(y+3)`

`y=1/8`

The given parabola intersects the y-axis at `(0, 1/8)` &

y-intercept: `=1/8`

Similarly, setting `y=0` in above equation of parabola, we get

`(x-5)^2=8(0+3)`

`x-5=\pm\sqrt24`

`x=5\pm2\sqrt6`

The given parabola intersects the x-axis at two points

`(5\pm2\sqrt6, 0)`

x-intercepts : `=5\pm2\sqrt6`

Answer 3

The vertex is `(5,-3)`.
The y-intercept is `(0,1/8)` or `(0,0.125)`.
The x-intercepts are `(5+2sqrt6,0)` and `(5-2sqrt6,0)`.
The approximate x-intercepts are `(9.899,0)` and `(0.101,0)`.

Explanation:

`y=1/8(x-5)^2-3` is a quadratic equation in vertex form:

`y=a(x-h)^2+k`,

where:

`h=5` and `k=-3`

The vertex is `(h,k)`, which is `(5,-3)`.

To find the y-intercept, substitute `0` for `x` and solve for `y`.

`y=1/8(0-5)^2-3`

`y=1/8(25)-3`

`y=25/8-3`

Simplify `3` to `24/8`.

`y=25/8-24/8=1/8=0.125`

The y-intercept is `(0,1/8)` or `(0,0.125)`.

To find the x-intercepts, substitute `0` for `y` and solve.

`0=1/8(x-5)^2-3`

Multiply both sides by `8`.

`0=(x-5)^2-24`

Expand `(x-5)^2`.

`0=x^2-10x+25-24`

`0=x^2-10x+1`,

where:

`a=1`, `b=-10`, `c=1`

Use the quadratic formula.

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

Plug in the known values and solve.

`x=(-(-10)+-sqrt((-10)^2-4*1*1))/(2*1)`

`x=(10+-sqrt(96))/2`

Prime factorize `96`.

`x=(10+-sqrt(2xx2xx2xx2xx2xx3))/2`

`x=(10+-sqrt(2^2xx2^2xx2xx3))/2`

`x=(10+-2xx2sqrt(2xx3))/2`

`x=(10+-4sqrt6)/2`

Factor out the common `2`.

`x=5+-2sqrt6`

`x=5+2sqrt6`, `5-2sqrt6`

Approximate values for `x`:

`x~~9.899, 0.101`

The x-intercepts are `(5+2sqrt6,0)` and `(5-2sqrt6,0)`.

The approximate x-intercepts are `(9.899,0)` and `(0.101,0)`.

graph{y=1/8(x-5)^2-3 [-4.29, 15.71, -4.64, 5.36]}