Explanation:
9.
we get the extreme values of y (a function) by finding the zeroes of the first derivative (that gives us the x-positions) and then using these x values in the original function.
a quadratic function has only one extreme value (1 less than the degree of the polynomial). but 2 zeroes (exactly the degree of the polynomial) - don't mix these two things up.
as the leading coefficient is positive (1), we know that the parabola is opening upwards, and the extreme value will be the vertex and therefore the minimum.
y = (x + 4)² + 6
y = x² + 8x + 16 + 6 = x² + 8x + 22
y' = 2x + 8
2x + 8 = 0
2x = -8
x = -4
(-4 + 4)² + 6 = 0² + 6 = 6
the smallest possible value of y = 6 (at x = -4).
10.
the intercept form of a parabola equation is
y = a(x - p)(x - q)
p and q are the x-intercept values.
in our case we see p = -3, q = 3.
also, we see the vertex being (0, 9), which we need to get "a".
y = a(x - -3)(x - 3) = a(x + 3)(x - 3)
using the vertex coordinates to get a :
9 = a(0 + 3)(0 - 3) = -9a
a = 9/-9 = -1
so, the quadratic equation is
y = -(x + 3)(x - 3) = -(x² - 9) = -x² + 9
11.
I assume the thinking here is : the flight curve of every falling object is a parabola.
the vertex of this parabola is (0, 36), as x represents the time in seconds, and y represents the height in ft.
the vertex form of a parabola is
y = a(x - h)² + k
with (h, k) being the vertex.
y = a(x - 0)² + 36 = ax² + 36
a must be negative, as the fall parabola opens downwards.
for typical fall situations on Earth a = -g/2 ≈ -16 (g being the gravity constant of 32.17405... ft/s²).
so, we have
y = -16x² + 36
for what x is y = 0 ?
0 = -16x² + 36
16x² = 36
x² = 36/16 = 9/4
x = 3/2 = 1.5
the coin hits the water after 1.5 seconds.