Question

PLS i need help !

Rewrite the equation in vertex form and identify the vertex, the axis of symmetry, and the y-intercept. y=-2x^2+12x-26

Answer 1

The equation of the polynomial in vertex form is
`y +8= (-2)\cdot (x-3)^(2)`, its vertex is
`(h,k) = (3, -8)`.

The expression of the axis of symmetry is
`x = 3`.

The y-intercept of the function is -26.

Explanation:

The vertex form of the second order polynomial is defined by the following expression:

`y-k = C\cdot (x-h)^(2)` (1)

Where:

`x` - Independent variable, dimensionless.

`y` - Dependent variable, dimensionless.

`h,k` - Coordinates of the vertex, dimensionless.

`C` - Vertex constant, dimensionless.

Let
`y = -2\cdot x^(2)+12\cdot x - 26`, then we proceed to present the produre for the determination of the vertex form:

1)
`y = -2\cdot x^(2)+12\cdot x - 26` Given

2)
`y = (-1)\cdot (2\cdot x^(2))+(-1)\cdot (-12\cdot x) + (-1)\cdot (26)`
`(-a)\cdot (-b) = a\cdot b`/
`(-a)\cdot b = -a\cdot b`

3)
`y = (-1)\cdot (2\cdot x^(2)-12\cdot x +26)` Distributive property

4)
`y = [(-1)\cdot (2)]\cdot (x^(2)-6\cdot x +13)` Associative and distributive properties

5)
`y = (-2)\cdot [(x^(2)-6\cdot x+9)+4]`
`(-a)\cdot b = -a\cdot b`

6)
`y = (-2) \cdot [(x-3)^(2)+4]` Perfect square trinomial

7)
`y = (-2)\cdot (x-3)^(2)+4\cdot (-2)` Distributive property

8)
`y = (-2)\cdot (x-3)^(2)+(-8)`
`(-a)\cdot b = -a\cdot b`

9)
`y +8= (-2)\cdot (x-3)^(2)` Compatibility of addition/Existence of the additive inverse/Modulative property/Result.

The equation of the polynomial in vertex form is
`y +8= (-2)\cdot (x-3)^(2)`, its vertex is
`(h,k) = (3, -8)`.

The axis of symmetry is a line perpendicular to axis in which the square component of the vertex form is set. The expression of the axis of symmetry is
`x = 3`.

The y-intercept is the value of the polynomial when
`x = 0`, then, the value is:

`y = -2\cdot (0)^(2)+12\cdot (0) -26`

`y = -26`

The y-intercept of the function is -26.