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

Answer:

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.

Step-by-step 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.