Step-by-step explanation:
1. a and b look correct.
you often write r like y, so it is not clear.
I also don't know what you mean by y (r ?) = 2x + 0 (6 ?).
let's check quickly :
point A gives us
I. -8 = p(-2)³ + q(-2)² + r(-2) = -8p + 4q - 2r
point B gives us
II. -2 = p(1)³ + q(1)² + r(1) = p + q + r
point C gives us
III. 0 = p(2)³ + q(2)² + r(2) = 8p + 4q + 2r
when we add I. and III. we get
-8 = 8q
q = -1
that makes II.
-2 = p - 1 + r
-1 = p + r
r = -1 - p
and that gives us for III.
0 = 8p - 4 + 2(-1 - p) = 8p - 4 - 2 - 2p = 6p - 6
6p = 6
p = 1
and II. then gives us
-2 = 1 - 1 + r
r = -2
so, yes, you did this correctly.
2.
f(x) = 0
a quadratic equation
ax² + bx + c = 0
has the generic solution
x = (-b ± sqrt(b² - 4ac))/(2a)
since both solutions (roots) are equal, it means that the square root part (discriminate) must be 0.
a = 2k
b = -4k
c = 1
(-4k)² - 4×2k×1 = 0
16k² - 8k = 0
2k² - k = 0
k(2k - 1) = 0
k = 0 is not a valid solution.
so,
2k - 1 = 0
2k = 1
k = 1/2 = 0.5
so, 2a. you did correctly too.
f(x) = x² - 2x + 1
2b.
y = p
the vertex of the parabola is
x = -b/2a = -(-2)/(2×1) = 2/2 = 1
f(1) = 1² - 2×1 + 1 = 0
as a is positive, the parabola opens upwards.
so, the vertex is the minimum.
that means all y (p) values are larger or equal to the y-value of the vertex.
therefore :
p >= 0
you forgot the "or equal" part. for the rest you were correct.
