The graph shows a quadratic function from -10 to 10 on the x and y axes. Points on the graph include (-5, 5), (-3, -3), (-2, -4), (-1, -3), and (1, 5).A graph is shown with the x-axis labeled negative 10 to 10 in increments of 1 and the y-axis labeled negative 10 to 10 in increments of 1. A parabola is shown opening upward with vertex left parenthesis negative 2 comma negative 4 right parenthesis and connecting the points left parenthesis negative 4 comma 0 right parenthesis and left parenthesis 0 comma 0 right parenthesis.
How many real solutions does the function shown on the graph have?
From the graph, we can see that the parabola opens upward and crosses the x-axis at two points (at x = -4 and x = 0). Since the leading coefficient of the quadratic function is positive (since the parabola opens upward), this means that the function has a minimum value and does not touch or cross the x-axis between these two points.
Therefore, the quadratic function has 2 real solutions.
During a softball game, a batter hits a ball upward from an initial height of 3 feet. The height, in feet, of the softball is given by s (t )= -16 t ^2 + 70 t + 3, where t is time in seconds and t is greater than or equal to 0. Which is closest to the time when the softball will be 50 feet above the ground?
To find the time when the softball will be 50 feet above the ground, we need to set the equation equal to 50 and solve for t:
-16t^2 + 70t + 3 = 50
Rearranging the equation:
-16t^2 + 70t - 47 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = -16, b = 70, and c = -47.
Plugging in these values, we get:
t = (-70 ± √(70^2 - 4(-16)(-47))) / (2(-16))
Simplifying further:
t = (-70 ± √(4900 - 2992)) / (-32)
t = (-70 ± √(1908)) / (-32)
t = (-70 ± 43.7) / (-32)
Now we can calculate the two possible values for t:
t1 = (-70 + 43.7) / (-32) ≈ 1.64
t2 = (-70 - 43.7) / (-32) ≈ 3.481
The closest time when the softball will be 50 feet above the ground is approximately 1.64 seconds.
To determine the number of real solutions that the quadratic function shown on the graph has, we need to examine the graph and identify the vertex.
From the information provided, we know that the vertex of the quadratic function is (-2, -4) since the parabola on the graph opens upward and connects the points (-4, 0) and (0, 0).
For a quadratic function in the form f(x) = ax^2 + bx + c, the number of real solutions can be determined by looking at the discriminant, which is given by the formula Δ = b^2 - 4ac.
In this case, since the parabola opens upward (a > 0), there are two possibilities for the number of real solutions:
1. If the discriminant is zero (Δ = 0), then there is only one real solution. This occurs when the vertex lies on the x-axis, and the parabola just touches it at one point.
2. If the discriminant is greater than zero (Δ > 0), then there are two distinct real solutions. This occurs when the parabola intersects the x-axis at two distinct points.
To find the discriminant, we can use the given points on the graph. Let's consider the first point, (-5, 5), and substitute it into the quadratic function:
5 = a(-5)^2 + b(-5) + c
Next, let's substitute the other two points, (-3, -3) and (1, 5), into the quadratic function:
-3 = a(-3)^2 + b(-3) + c
5 = a(1)^2 + b(1) + c
We now have a system of three equations with three variables (a, b, c). By solving this system, we can determine the values of a, b, and c - and hence find the quadratic function. From there, we can calculate the discriminant Δ and determine the number of real solutions.
Unfortunately, the given points on the graph do not provide a sufficient number of equations to solve for a, b, and c accurately. To determine the number of real solutions, we would need additional information or additional points on the graph.