Answer: [tex]a)\ \text{Vertex}:y=-\dfrac{3}{2}(x+1)^2+6[/tex]
[tex]b)\ \text{Standard}:y=-\dfrac{3}{2}x^2-3x=\dfrac{9}{2}[/tex]
c) Transformations: reflection over the x-axis,
vertical stretch by a factor of 3/2,
horizontal shift 1 unit to the left,
vertical shift 6 units up
Step-by-step explanation:
Intercept form: y = a(x - p)(x - q)
Vertex form: y = a(x - h)² + k
Standard form: y = ax² + bx + c
We can see that the new vertex is (-1, 6). Use the Intercept form to find the vertical stretch: y = a(x - p)(x - q) where p, q are the intercepts.
p = -3, q = 1, (x, y) = (-1, 6)
a(-1 + 3)(-1 -1) = 6
a (2)(-2) = 6
a = -6/4
a = -3/2
a) Input a = -3/2 and vertex (h, k) = (-1, 6) into the Vertex form to get:
[tex]y=-\dfrac{3}{2}(x+1)^2+6[/tex]
b) Input a = -3/2 into the Intercept form and expand to get the Standard form:
[tex]y=-\dfrac{3}{2}(x+3)(x-1)\\\\\\y=-\dfrac{3}{2}(x^2+2x-3)\\\\\\y=-\dfrac{3}{2}x^2-3x+\dfrac{9}{2}[/tex]
c) Use the Vertex form to identify the transformations:
[tex]y=-\dfrac{3}{2}(x+1)^2+6[/tex]
- a is negative: reflection over the x-axis
- |a| = 3/2: vertical stretch by a factor of 3/2
- h = -1: horizontal shift left 1 unit
- k = +6: vertical shift up 6 units
