a parabola has an y-intercept at y=11 and a vertex at (4,3)
a)find the other x-intercept
b)find another point on the parabola
a) To find the x-intercept of a parabola, we set y equal to zero and solve for x. Since the y-intercept is 11, we know that the point (0,11) lies on the parabola.
To find the x-intercept, we set y equal to zero and solve the equation:
0 = a(x - 4)^2 + 3
Plugging in the coordinates of the y-intercept (0, 11) gives us:
0 = a(0 - 4)^2 + 3
0 = 16a + 3
Solving for a gives us:
16a = -3
a = -3/16
Now we can plug this value of a back into the equation and solve for x:
0 = (-3/16)(x - 4)^2 + 3
(-3/16)(x - 4)^2 = -3
(x - 4)^2 = 3 * (-16/3)
(x - 4)^2 = -16
Taking the square root of both sides, we get:
x - 4 = ± √(-16)
x - 4 = ± 4i
x = 4 ± 4i
Therefore, the parabola does not have any x-intercepts in the real number system.
b) To find another point on the parabola, we can use the equation y = a(x - 4)^2 + 3. Let's plug in a value for x and solve for y.
Let's choose x = 2:
y = (-3/16)(2 - 4)^2 + 3
y = (-3/16)(-2)^2 + 3
y = (-3/16)(4) + 3
y = -3/4 + 3
y = -3/4 + 12/4
y = 9/4
Therefore, another point on the parabola is (2, 9/4).