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11Alexandr11 [23.1K]
3 years ago
15

Please walk me through how to do this so I can di the other questions​

Mathematics
1 answer:
Naddik [55]3 years ago
3 0

Answer:

Axis is a vertical line at x = 2

Vertex is (2, -1)

y-intercept is (0, 3)

Solutions are x = 1 and x = 3

Step-by-step explanation:

To draw the graph of the quadratic equation you must find at least 5 points lie on the graph by choose values of x and find their values of y

Let us do that

Use x = -1, 0, 1, 2, 3, 4, 5

∵ y = x² - 4x + 3

∵ x = -1

∴ y = (-1)² - 4(-1) + 3 = 1 + 4 + 3 = 8

→ Plot point (-1, 8)

∵ x = 0

∴ y = (0)² - 4(0) + 3 = 0 + 0 + 3 = 3

→ Plot point (0, 3)

∵ x = 1

∴ y = (1)² - 4(1) + 3 = 1 - 4 + 3 = 0

→ Plot point (1, 0)

∵ x = 2

∴ y = (2)² - 4(2) + 3 = 4 - 8 + 3 = -1

→ Plot point (2, -1)

∵ x = 3

∴ y = (3)² - 4(3) + 3 = 9 - 12 + 3 = 0

→ Plot point (3, 0)

∵ x = 4

∴ y = (4)² - 4(4) + 3 = 16 - 16 + 3 = 3

→ Plot point (4, 3)

∵ x = 5

∴ y = (5)² - 4(5) + 3 = 25 - 20 + 3 = 8

→ Plot point (5, 8)

→ Join all the points to form the parabola

From the graph

∵ The axis of symmetry is the vertical line passes through the vertex point

∵ x-coordinate of the vertex point is 2

∴ Axis is a vertical line at x = 2

∵ The coordinates of the vertex point of the parabola are (2, -1)

∴ Vertex is (2, -1)

∵ The parabola intersects the y-axis at point (0, 3)

∴ y-intercept is (0, 3)

∵ x² - 4x + 3 = 0

∵ The solutions of the equation are the values of x at y = 0

→ That means the intersection points of the parabola and the x-axis

∵ The parabola intersects the x-axis at points (1, 0) and (3, 0)

∴ Solutions are x = 1 and x = 3

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After eliminating radicals, what quadratic equation can you solve to find the potential solutions of sqrt 2x+3 - sqrt x+1 = 1
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Answer:

Step-by-step explanation:

We have given:

√2x+3 - √x+1  = 1

First of all isolate the square root of the left hand side:

√2x+3 = √x+1 +1

Now take square on both sides.

(√2x+3)^2 = (√x+1 +1)^2

Open the R.H.S by squaring formula.

∴(a+b)^2 = a^2+2ab+b^2

2x+3 = (√x+1)^2 + 2(√x+1)(1)+(1)^2

2x+3 = x+1 +2√x+1 +1

2x+3 = x+2 +2√x+1

Combine the like terms:

2x-x+3-2 = 2√x+1

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Take square on both sides

(x+1)^2 = (2√x+1)^2

x²+2x+1 = 4x+4

x²+2x-4x+1-4 = 0

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Now solve the quadratic equation:

a = 1 , b= -2 , c = -3

x = -b+/-√b²-4ac/2a

x = -(-2)+/-√(-2)² - 4(1)(-3) / 2(1)

x = 2 +/- √4+12 / 2

x = 2+/- √16/2

x = 2+/- 4 /2

x = 2+4/2     , x = 2-4/2

x = 6/2      , x = -2/2

x = 3      , x = -1

The solutions we get is (3, -1).

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