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.
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?
3 answers
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.
-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.