Question

2v^2-12v-29=3 Find the Vertex Form and Vertex

Answer 1

Answer:

see explanation

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

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

given a quadratic in standard form : y = ax² + bx + c  : a ≠ 0

Then the x- coordinate of the vertex is

$x_{vertex}$ = - $\frac{b}{2a}$

Rearrange 2v² - 12v - 29 = 3 into standard form ( subtract 3 from both sides )

2v² - 12v - 32 = 0 ← in standard form

with a = 2, b = - 12, hence

$x_{vertex}$ = - $\frac{-12}{4}$ = 3

substitute x = 3 into the equation for corresponding value of y

y = 2(3)² - 12(3) - 32 = 18 - 36 - 32 = - 50

vertex = (3, - 50)

y = 2(v- 3)² - 50 ← in vertex form

Answer 2

Answer:

The vertex form is $2(v-3)^{2}-50$

The vertex is (3 , -50)

Step-by-step explanation:

∵ $2v^{2}-12v-29 = 3$ ⇒ $2v^{2}-12v-29-3=2v^{2}-12v-32=a(v+b)^{2}+c$, where a , b and c are constant

$2v^{2}-12v-32=a(v^{2}+2bv+b^{2})+c$

$2v^{2}-12v-32=av^{2}+2abv+ab^{2}+c$⇒ compare the two sides

a = 2

2ab = -12 ⇒ 2(2)b = -12 ⇒ 4b = -12 ⇒ b = -3

$ab^{2}+c=-32$ ⇒ $2(-3)^{2}+c=-32$

2 × 9 + c = -32 ⇒ c = -32 -18 = -50

∴ The vertex form is $2(v-3)^{2}-50$

∴ The vertex is (3 , -50)