Math / Algebra

Question3LEM 9: THE HAWK PROBLEM A hawk is on top of a tree and spots a fish in the water. The hawks's pathway in metres is measured by time in seconds by the equation: $y=x^{2}-6 x+11$ a) When does the hawk enter the water? b) Explain why this is a problem? c) Sketch a graph

Studdy Solution

STEP 1

1. The equation $y = x^2 - 6x + 11$ describes the hawk's pathway.
2. The hawk enters the water when $y = 0$ .
3. We need to find the time $x$ when $y = 0$ .
4. The problem may involve the nature of the solution or the context of the scenario.
5. A sketch of the graph will help visualize the pathway.

STEP 2

1. Solve for when the hawk enters the water.
2. Analyze the problem for potential issues.
3. Sketch the graph of the pathway.

STEP 3

To find when the hawk enters the water, set the equation $y = x^2 - 6x + 11$ equal to zero and solve for $x$ .
$x^2 - 6x + 11 = 0$

STEP 4

Use the quadratic formula to solve for $x$ :
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$ , $b = -6$ , and $c = 11$ .

STEP 5

Calculate the discriminant $b^2 - 4ac$ :
$(-6)^2 - 4 \times 1 \times 11 = 36 - 44 = -8$
Since the discriminant is negative, there are no real solutions.

STEP 6

Explain why this is a problem: The negative discriminant indicates that the hawk's pathway does not intersect the water level (y=0), meaning the hawk never enters the water.

STEP 7

To sketch the graph, note that the equation $y = x^2 - 6x + 11$ is a parabola opening upwards with vertex at $x = \frac{-b}{2a} = \frac{6}{2} = 3$ .

STEP 8

Calculate the vertex $y$ -value:
$y = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2$
The vertex is at $(3, 2)$ .

STEP 9

The parabola does not intersect the x-axis, confirming it does not reach $y = 0$ . Sketch the graph with the vertex at $(3, 2)$ and opening upwards.
The hawk never enters the water as the pathway does not reach $y = 0$ . The graph is a parabola with vertex at $(3, 2)$ .

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Question3LEM 9: THE HAWK PROBLEM A hawk is on top of a tree and spots a fish in the water. The hawks's pathway in metres is measured by time in seconds by the equation: $y=x^{2}-6 x+11$ a) When does the hawk enter the water? b) Explain why this is a problem? c) Sketch a graph

Studdy Solution

STEP 1

1. The equation $y = x^2 - 6x + 11$ describes the hawk's pathway.
2. The hawk enters the water when $y = 0$ .
3. We need to find the time $x$ when $y = 0$ .
4. The problem may involve the nature of the solution or the context of the scenario.
5. A sketch of the graph will help visualize the pathway.

STEP 2

1. Solve for when the hawk enters the water.
2. Analyze the problem for potential issues.
3. Sketch the graph of the pathway.

STEP 3

To find when the hawk enters the water, set the equation $y = x^2 - 6x + 11$ equal to zero and solve for $x$ .
$x^2 - 6x + 11 = 0$

STEP 4

Use the quadratic formula to solve for $x$ :
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$ , $b = -6$ , and $c = 11$ .

STEP 5

Calculate the discriminant $b^2 - 4ac$ :
$(-6)^2 - 4 \times 1 \times 11 = 36 - 44 = -8$
Since the discriminant is negative, there are no real solutions.

STEP 6

Explain why this is a problem: The negative discriminant indicates that the hawk's pathway does not intersect the water level (y=0), meaning the hawk never enters the water.

STEP 7

To sketch the graph, note that the equation $y = x^2 - 6x + 11$ is a parabola opening upwards with vertex at $x = \frac{-b}{2a} = \frac{6}{2} = 3$ .

STEP 8

Calculate the vertex $y$ -value:
$y = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2$
The vertex is at $(3, 2)$ .

STEP 9

The parabola does not intersect the x-axis, confirming it does not reach $y = 0$ . Sketch the graph with the vertex at $(3, 2)$ and opening upwards.
The hawk never enters the water as the pathway does not reach $y = 0$ . The graph is a parabola with vertex at $(3, 2)$ .

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Question3LEM 9: THE HAWK PROBLEM A hawk is on top of a tree and spots a fish in the water. The hawks's pathway in metres is measured by time in seconds by the equation: y=x2−6x+11y=x^{2}-6 x+11y=x2−6x+11 a) When does the hawk enter the water? b) Explain why this is a problem? c) Sketch a graph

Studdy Solution

STEP 1

STEP 2

1. Solve for when the hawk enters the water.2. Analyze the problem for potential issues.3. Sketch the graph of the pathway.

STEP 3

To find when the hawk enters the water, set the equation y=x2−6x+11 y = x^2 - 6x + 11 y=x2−6x+11 equal to zero and solve for x x x.x2−6x+11=0 x^2 - 6x + 11 = 0 x2−6x+11=0

STEP 4

Use the quadratic formula to solve for x x x:x=−b±b2−4ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} x=2a−b±b2−4ac​​where a=1 a = 1 a=1, b=−6 b = -6 b=−6, and c=11 c = 11 c=11.

STEP 5

Calculate the discriminant b2−4ac b^2 - 4ac b2−4ac:(−6)2−4×1×11=36−44=−8 (-6)^2 - 4 \times 1 \times 11 = 36 - 44 = -8 (−6)2−4×1×11=36−44=−8Since the discriminant is negative, there are no real solutions.

STEP 6

Explain why this is a problem: The negative discriminant indicates that the hawk's pathway does not intersect the water level (y=0), meaning the hawk never enters the water.

STEP 7

To sketch the graph, note that the equation y=x2−6x+11 y = x^2 - 6x + 11 y=x2−6x+11 is a parabola opening upwards with vertex at x=−b2a=62=3 x = \frac{-b}{2a} = \frac{6}{2} = 3 x=2a−b​=26​=3.

STEP 8

Calculate the vertex y y y-value:y=(3)2−6(3)+11=9−18+11=2 y = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2 y=(3)2−6(3)+11=9−18+11=2The vertex is at (3,2) (3, 2) (3,2).

STEP 9

The parabola does not intersect the x-axis, confirming it does not reach y=0 y = 0 y=0. Sketch the graph with the vertex at (3,2) (3, 2) (3,2) and opening upwards. The hawk never enters the water as the pathway does not reach y=0 y = 0 y=0. The graph is a parabola with vertex at (3,2) (3, 2) (3,2).

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Question3LEM 9: THE HAWK PROBLEM A hawk is on top of a tree and spots a fish in the water. The hawks's pathway in metres is measured by time in seconds by the equation: $y=x^{2}-6 x+11$ a) When does the hawk enter the water? b) Explain why this is a problem? c) Sketch a graph

1. Solve for when the hawk enters the water.
2. Analyze the problem for potential issues.
3. Sketch the graph of the pathway.

To find when the hawk enters the water, set the equation $y = x^2 - 6x + 11$ equal to zero and solve for $x$ .
$x^2 - 6x + 11 = 0$

Use the quadratic formula to solve for $x$ :
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
where $a = 1$ , $b = -6$ , and $c = 11$ .

Calculate the discriminant $b^2 - 4ac$ :
$(-6)^2 - 4 \times 1 \times 11 = 36 - 44 = -8$
Since the discriminant is negative, there are no real solutions.

To sketch the graph, note that the equation $y = x^2 - 6x + 11$ is a parabola opening upwards with vertex at $x = \frac{-b}{2a} = \frac{6}{2} = 3$ .

Calculate the vertex $y$ -value:
$y = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2$
The vertex is at $(3, 2)$ .

The parabola does not intersect the x-axis, confirming it does not reach $y = 0$ . Sketch the graph with the vertex at $(3, 2)$ and opening upwards.
The hawk never enters the water as the pathway does not reach $y = 0$ . The graph is a parabola with vertex at $(3, 2)$ .