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For the parabola y = 2x² + 3x + 1,
a. i. The x-intercepts are x = -1/2 and x = -1
a. ii. The y-intercept is y = 1
b. i. The line of symmetry is at x = -3/4
b. ii. The equation of the line of symmetry is x = -3/4
c. The vertex is at (-3/4, -1/8)
A parabola is part of a conic section which is a curve.
Given the parabola y = 2x² + 3x + 1, we find the x- intercept when y = 0.
So, y = 2x² + 3x + 1
2x² + 3x + 1 = 0
2x² + 2x + x + 1 = 0
Factorizing, we have
2x(x + 1) + (x + 1) = 0
(2x + 1)(x + 1) = 0
2x + 1 = 0 or x + 1 = 0
2x = -1 or x = -1
x = -1/2 or x = -1
So, the x-intercepts are x = -1/2 and x = -1
Given the parabola y = 2x² + 3x + 1, we find the y- intercept when x = 0.
So, y = 2x² + 3x + 1
So, y = 2(0)² + 3(0) + 1
y = 0 + 0 + 1
y = 1
So, the y-intercept is y = 1
The line of symmetry passes through the vertex of the parabola where dy/dx = 0.
So, y = 2x² + 3x + 1
dy/dx = d(2x² + 3x + 1)/dx
= d(2x²)/dx + d3x/dx + d1/dx
= 4x + 3 + 0
= 4x + 3
So, when dy/dx = 0,we have
4x + 3 = 0
4x = -3
x = -3/4
So, the line of symmetry is at x = -3/4
The equation of the line of symmetry is x = -3/4
Since the vertex of the parabola is a t x = -3/4, substituting this into the equation, we have
y = 2x² + 3x + 1
y = 2(-3/4)² + 3(-3/4) + 1
y = 2(9/16) - 9/4 + 1
y = 9/8 - 9/4 + 1
y = (9 - 18 + 8)/8
y = (17 - 18)/8
y = -1/8
So, the vertex is at (-3/4, -1/8)
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