I really need help on some sample work. Just need to check my answers:
1. Describe how the graph of y=-x^2+1 differs from y=x^2. Then find the axis of symmetry and the vertex
2. Graph the quadratic equation y=x^2+x-2. Identify the axis of symmetry and the vertex
3. A ball is thrown from the top of a 50-ft building with an upward velocity of 24 ft/s. When will it reach its maximum height? How far above the ground will it be? Use the equation
h(t)=-16+24+50
4. Solve the quadratic equation 3x^2=27
5. Solve the quadratic equation 2x^2-4x+1=0
6. How many solutions are in the equation
3x^2+4x-7=0
7. Solve by completing the square: A rectangular patio has a length of (x+6) m, a width of (x+8) m, and a total area of 400 m^2. Find the dimensions to the nearest tenth.
8. Find the radius of a circular lid with an area of 12in.^2
9. Solve the system of equations
{ y=x^2+9
y=6x
10. Solve the system of equations
{ y=4x+8
x^2+7x=20
What, no ideas on any of them?
yeah - sample work - >wink wink<
Sure! I can help you with your sample work. Let's go question by question:
1. To describe how the graph of y=-x^2+1 differs from y=x^2, we can compare their shapes and orientations. The graph of y=x^2 is a U-shaped curve that opens upwards, while the graph of y=-x^2+1 is also a U-shaped curve, but it opens downwards. The negative sign in front of x^2 in the second equation reflects the graph vertically.
To find the axis of symmetry of both equations, we use the formula x = -b/2a. For y=x^2, the coefficient of x is 0, so the axis of symmetry is x = 0. For y=-x^2+1, a = -1 and b = 0, so the axis of symmetry is also x = 0.
The vertex of a quadratic equation can be found by substituting the value of x from the axis of symmetry into the equation. For y=x^2, when x = 0, y = 0^2 = 0. So the vertex is (0, 0). Similarly, for y=-x^2+1, when x = 0, y = -(0^2) + 1 = 1. So the vertex is (0, 1).
2. To graph the quadratic equation y=x^2+x-2, we can plot some points and draw a curve through them. Let's start by substituting some x values into the equation. For example, when x = -2, y = (-2)^2 + (-2) - 2 = 0. So we have one point (-2, 0).
Similarly, when x = -1, y = (-1)^2 + (-1) - 2 = -2. Another point is (-1, -2).
Using more x values, you can find corresponding y values and plot more points. Once you have enough points, connect them to form a smooth curve.
To find the axis of symmetry, we use the formula x = -b/2a. For y=x^2+x-2, a = 1 and b = 1. Thus, x = -1/2 is the axis of symmetry.
To find the vertex, substitute the value of x from the axis of symmetry into the equation. When x = -1/2, y = (-1/2)^2 + (-1/2) - 2 = -9/4. Therefore, the vertex is (-1/2, -9/4).
3. To find when the ball reaches its maximum height and how far above the ground it will be, we use the given equation h(t) = -16t^2 + 24t + 50. This equation represents the height of the ball at time t.
To find when the ball reaches its maximum height, we know that the vertex of a quadratic equation occurs at t = -b/2a. In this equation, a = -16 and b = 24. So, t = -24/(2*-16) = -24/-32 = 3/4. Therefore, the ball reaches its maximum height after 3/4 seconds.
To find how far above the ground the ball will be at its maximum height, substitute the value of t into the equation. When t = 3/4, h(3/4) = -16*(3/4)^2 + 24*(3/4) + 50 = 44 feet. Thus, the ball will be 44 feet above the ground at its maximum height.
I will continue explaining how to answer the rest of the questions in subsequent responses.