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Y=2x^2 + 3x+1
a.Find the x - and y-intercept
b. Where is the line of symmetry of this parábola? Write its equation
c. Find the coordinates of the vertex

Sagot :

oladayoademosu

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)

What is a parabola?

A parabola is part of a conic section which is a curve.

a. i. Find the x - intercept

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

ii Find the y intercept?

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

b. i. Where is the line of symmetry of this parábola?

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

ii. Write its equation

The equation of the line of symmetry is x = -3/4

c. Find the coordinates of the vertex?

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)

Learn more about parabola here:

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