# Determine the Vertex and the x intercepts for the function: y = -2(x - 1)^2 + 24 a) v(1, 24); 13, -11 b) v(1, -24); 13, -11 c) v(1, 24); none d) v(1, 24); 1+ 2sqrt(3), 1 - 2sqrt(3) e)...

Determine the Vertex and the x intercepts for the function:

y = -2(x - 1)^2 + 24

a) v(1, 24); 13, -11

b) v(1, -24); 13, -11

c) v(1, 24); none

d) v(1, 24); 1+ 2sqrt(3), 1 - 2sqrt(3)

e) v(1, -24); 1 + 2sqrt(3), 1 - 2sqrt(3)

f) v(1, -24); none

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### 2 Answers

Determine the vertex and the x-intercepts for the function

`y=-2(x-1)^2+24`

This is in vertex form `y=a(x-h)^2+k` where the vertex is at (h,k). Here the vertex is (1,24).

To find the x-intercepts we set y=0:

`0=-2(x-1)^2+24`

`2(x-1)^2=24`

`(x-1)^2=12`

`x-1=+-sqrt(12)`

`x-1=+-2sqrt(3)`

`x=1+-2sqrt(3)`

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The vertex is at (1,24) and the intercepts are `x=1+-2sqrt(3)` so the answer is (d)

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Write the equation `y=-2(x-1)^2+24` in standard form.

`y=-2(x^2-2x+1)+24`

`y=-2x^2+4x-2+24`

`y=-2x^2+4x+22`

The x coordinate of the vertex `x_v` of a parabola `y=ax^2+bx+c` is:

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

Substitute 4 for b and -2 for a.

`x_v=(-4)/((2)(-2))=1`

Determine y(1)

`y=-2(1)^2+4(1)+22`

`y=-2+4+22=24`

**Thus the vertex is at (1,24)**

Determine the x intercepts by equating y to 0.

`-2x^2+4x+22=0`

` ` Divide by -2 to simplify.

`x^2-2x-11=0`

Use the quadratic formula to solve.

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

Substitute -2 for b, 1 for a and -11 for c.

`x=(2+-sqrt((-2)^2-(4)(1)(-11)))/((2)(1))`

`x=(2+-sqrt(48))/2`

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

`x=1+2sqrt3`

`x=1-2sqrt3`

**Thus the x intercepts are `1+2sqrt3` and `1-2sqrt3` **

**Thus the answer is d)**