Solving Algebra 10 page 49 textbook

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Solving Algebra 10 page 49 textbook - Quadratic Functions

Main focus today is on exercises related to quadratic functions, aimed to guide students through Algebra 10 page 49 textbook. Let's delve into the details of this lesson.

=> For more 10th grade math problem-solving, check here: Solving 10th Grade Math

Solving Algebra 10 page 49 textbook - Quadratic Functions

Exercise 1 (Math 10 Textbook Page 49)
Identify the vertex coordinates and intersection points with the x-axis, y-axis (if any) of a parabola:
a) y = x² - 3x + 2
b) y = -2x² + 4x - 3
c) y = x² - 2x
d) y = -x² + 4
Solution:
a) Vertex at (1, -1), Intersects y-axis at (0, 2), Intersects x-axis at (1, 0) and (2, 0)
b) Vertex at (1, -1), Intersects y-axis at (0, -3), No intersection with x-axis.
c) Vertex at (1, -1), Intersects y-axis at (0, 0), Intersects x-axis at (0, 0) and (2, 0)
d) Vertex at (0, 4), Intersects y-axis at (0, 4), Intersects x-axis at (-2, 0) and (2, 0)
Exercise 2 (Math 10 Textbook Page 49)
Construct the table of variations and draw the graph of the following functions:
a) y = 3x² - 4x + 1
b) y = -3x² + 2x -1
c) y = 4x² - 4x + 1
d) y = -x² + 4x - 4
e) y = 2x² + x + 1
f) y = -x² + x - 1

Exercise 3 (Math 10 Textbook Page 49)
Determine the parabola y = ax² + bx + 2, knowing that the parabola:
a) Passes through two points M(1,5) and N(-2,8)
b) Passes through two points A(3,-4) and has the axis of symmetry x = -3/2
c) Has vertex at I(2,-2)
d) Passes through point B(-1,6) and the ordinate of the vertex is -1/4
Solution:
a) The parabola equation is: y = 2x² + x + 2
b) The parabola equation is: y = -1/3x² - x + 2
c) The parabola equation is: y = x² - 4x + 4
d) The parabola equation is: y = -2x² + 3x + 5

Exercise 4 (Math 10 Textbook Page 49)
Determine the parabola y = ax² + bx + 2, knowing that the parabola:
a) Passes through two points M(1,5) and N(-2,8)
b) Passes through two points A(3,-4) and has the axis of symmetry x = -3/2
c) Has vertex at I(2,-2)
d) Passes through point B(-1,6) and the ordinate of the vertex is -1/4
Solution:
a) The parabola equation is: y = 2x² + x + 2
b) The parabola equation is: y = -1/3x² - x + 2
c) The parabola equation is: y = x² - 4x + 4
d) The parabola equation is: y = -2x² + 3x + 5

Exercise 5 (Math 10 Textbook Page 49)
Determine the parabola y = ax² + bx + 2, knowing that the parabola:
a) Passes through two points M(1,5) and N(-2,8)
b) Passes through two points A(3,-4) and has the axis of symmetry x = -3/2
c) Has vertex at I(2,-2)
d) Passes through point B(-1,6) and the ordinate of the vertex is -1/4
Solution:
a) The parabola equation is: y = 2x² + x + 2
b) The parabola equation is: y = -1/3x² - x + 2
c) The parabola equation is: y = x² - 4x + 4
d) The parabola equation is: y = -2x² + 3x + 5

=> The parabola equation is: y = x² - 4x + 2
d) As the parabola passes through point B (-1;6) and the ordinate of the vertex is -1/4, we get:

=> The parabola equation is either y = x² - 3x + 2 or y = 16x² + 12x + 2
Exercise 4 (Math 10 Textbook Page 49)
Determine a, b, c knowing that the parabola y = ax² + bx + c passes through point A(8;0) and has vertex I(6;-12)
Solution:
Since y = ax² + bx + c passes through point A(8;0) and has vertex I(6;-12), we have:

=> The parabola equation is: y = 3x² - 36x + 96

The solution section for Exercise 1: Determine the coordinates of the vertex and the intersection points with the x-axis, y-axis of the parabolas.

Exercise 2: Constructing the variation table and plotting the function graphExercise 3: Determining the equation of a parabolaExercise 4: Finding the coefficients of the parabola equation given the vertex coordinates and passing through a given point

Solution for questions 1 to 4 on page 49 of Math textbook for grade 10 volume 1

- Solve question 1 on page 49 of Math textbook for grade 10 volume 1

- Solve question 2 on page 49 of Math textbook for grade 10 volume 1

- Solve question 3 on page 49 of Math textbook for grade 10 volume 1

- Solve question 4 on page 49 of Math textbook for grade 10 volume 1

Guidance for solving exercises on page 49 of Algebra 10 textbook in the exercise solving section for 10th-grade math. Students can review the solutions to exercises on page 45 of Geometry 10 textbook provided in the previous lesson or preview the guidance for solving exercises on pages 50, 51 of Algebra 10 textbook to improve their understanding of 10th-grade math.

Frequently Asked Questions

What are the vertex coordinates of the parabola y = x² - 3x + 2?

The vertex coordinates of the parabola defined by the equation y = x² - 3x + 2 are (1, -1). This indicates the point at which the parabola reaches its minimum value.

How can I find intersection points of a parabola with axes?

To find the intersection points of a parabola with the axes, substitute zero for y to find x-intercepts and zero for x to find the y-intercept. Analyze the resulting equations to identify points on the graph.

What is the method for constructing the variation table for quadratic functions?

To construct the variation table for a quadratic function, identify the vertex, determine intercepts, analyze the function's direction (concave up or down), and plot these key points to visualize how the function behaves across its domain.

How do I determine the equation of a parabola given specific points?

To determine the equation of a parabola given specific points, use the general form y = ax² + bx + c. Substitute the coordinates of the points into the equation to create a system of equations, then solve for a, b, and c.

What is the significance of the vertex in a parabola's graph?

The vertex of a parabola is significant because it represents the maximum or minimum point of the graph, depending on whether the parabola opens upwards or downwards. It also helps define the parabola's symmetry.

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