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How do I find the vertex and intercepts for the two quadratic functions listed below?

0 votes
How do I sketch and Graph them as well.
a. y=x^2+4x
b. y=-2x^2+8x
asked Mar 17, 2014 in ALGEBRA 1 by dkinz Apprentice

3 Answers

0 votes

(a).The equation is y = x2 + 4x.

Write the equation : y = x2 + 4x in complete square form.

To change the expressions (x2 + 4x) into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

 Here x coefficient = 4. so, (half the x coefficient)² = (4/2)2= 4.

Add 4 to each side of the equation.

y + 4 = x2 + 4x + 4

y + 4 = (x + 2)2

y = (x + 2)2 - 4.

The above equation represent the parabola.

Compare the equation y = (x + 1)2 - 4 with standard form of the equation of a parabola with vertex (h, k) and axis of symmetry x = h is y = a(x - h)2 + k.

Vertex (h, k ) = (- 2, - 4), and axis of symmetry x = h = - 2.

To find the y - intercept, substitute x = 0 in the y = (x + 1)2 - 4.

y = (0 + 2)2 - 4

y = 4 - 4

y = 0.

To find the x - intercept, substitute y = 0 in the y = (x + 1)2 - 4.

0 = (x + 2)2 - 4.

4 = (x + 2)2

± 2 = x + 2

 - 2 ± 2 = x

x = 0 and x = - 4.

answered Apr 3, 2014 by steve Scholar
0 votes

Continued :

Make the table of values to find ordered pairs that satisfy the equation.

Choose values for x and find the corresponding values for y.

x

(x, y)

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1. Draw a coordinate plane.

2. Plot the coordinate points.

3. Connect these points with smooth curve.

graph the equation y=(x+2)^2-4

The vertex = (h, k ) = (- 2, - 4), y - intercept = 0 and x - intercepts = 0 and - 4.

answered Apr 3, 2014 by steve Scholar
0 votes

(b). The equation is y = - 2x2 + 8x.

Write the equation : y = - 2x2 + 8x in complete square form.

y = - 2(x2 - 4x)

To change the expressions (x2 - 4x) into a perfect square trinomial add (half the x coefficient)² to each side of the expression.

 Here x coefficient = 4. so, (half the x coefficient)² = (4/2)2= 4.

Add - 2(4) to each side of the equation.

y - 2(4) = - 2(x2 - 4x + 4)

y - 8 = - 2(x - 2)2

y = - 2(x - 2)2 + 8.

The above equation represent the parabola.

Compare the equation y = - 2(x - 2)2 + 8 with standard form of the equation of a parabola with vertex (h, k) and axis of symmetry x = h is y = a(x - h)2 + k.

Vertex (h, k ) = (2, 8), and axis of symmetry x = h = 2.

To find the y - intercept, substitute x = 0 in the y = - 2(x - 2)2 + 8.

y = - 2(0 - 2)2 + 8

y = - 2(4) + 8

y = 0.

To find the x - intercept, substitute y = 0 in the y = - 2(x - 2)2 + 8.

0 = - 2(x - 2)2 + 8.

(- 8)/(-2) = (x - 2)2

4 = (x - 2)2

± 2 = x - 2

2 ± 2 = x

x = 4 and x = 0.

Make the table of values to find ordered pairs that satisfy the equation.

Choose values for x and find the corresponding values for y.

x

y = - 2x2 + 8x

(x, y)

-1 y = - 2(-1)2 + 8(-1) = - 2 - 8 = - 10 (-1, -10)

0

y = - 2(0)2 + 8(0) = 0

(0, 0)

0

y = - 2(1)2 + 8(1) = - 2 + 8 = 6

(1, 6)

2

y = - 2(2)2 + 8(2) = - 2(4) + 16 = - 8 + 16 = 8

(2, 8)

3

y = - 2(3)2 + 8(3) = - 2(9) + 24 = - 18 + 24 = 6

(3, 6)

4

y = - 2(4)2 + 8(4) = - 2(16) + 32 = - 32 + 32 = 0

(4, 0)

5 y = - 2(5)2 + 8(5) = - 2(25) + 40 = - 50 + 40 = - 10 (5, 10)

1. Draw a coordinate plane.

2. Plot the coordinate points.

3. Connect these points with a smooth curve.

graph the equation y=-2x^2+8x

The vertex = (h, k ) = (2, 8), y - intercept is 0 and x - intercepts are 0 and - 4.

answered Apr 3, 2014 by steve Scholar

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