Answer :
To analyze the quadratic equation [tex]\( y = -16x^2 + 111x + 50 \)[/tex], we will take the following steps:
Step 1: Determine the Roots
The roots of the quadratic equation can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In our equation, [tex]\( a = -16 \)[/tex], [tex]\( b = 111 \)[/tex], and [tex]\( c = 50 \)[/tex]. Plugging these values into the quadratic formula, we get:
[tex]\[ x = \frac{-111 \pm \sqrt{{111}^2 - 4(-16)(50)}}{2(-16)} \][/tex]
Calculating inside the square root first:
[tex]\[ b^2 - 4ac = 111^2 - 4(-16)(50) = 12321 + 3200 = 15521 \][/tex]
Then, the square root of 15521 is approximately:
[tex]\[ \sqrt{15521} \approx 124.599 \][/tex]
Now we compute the two potential solutions for [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{-111 + 124.599}{-32} \approx \frac{13.599}{-32} \approx -0.424 \][/tex]
[tex]\[ x_2 = \frac{-111 - 124.599}{-32} \approx \frac{-235.599}{-32} \approx 7.362 \][/tex]
So, the roots of the equation are approximately [tex]\( x_1 \approx -0.424 \)[/tex] and [tex]\( x_2 \approx 7.362 \)[/tex].
Step 2: Determine the Vertex
For a parabola expressed in the form [tex]\( y = ax^2 + bx + c \)[/tex], the x-coordinate of the vertex can be found using:
[tex]\[ x = \frac{-b}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = \frac{-111}{2(-16)} = \frac{-111}{-32} \approx 3.469 \][/tex]
To find the y-coordinate of the vertex, plug [tex]\( x = 3.469 \)[/tex] back into the original equation:
[tex]\[ y = -16(3.469)^2 + 111(3.469) + 50 \][/tex]
Calculating each component:
[tex]\[ -16(3.469)^2 \approx -16(12.034) \approx -192.544 \][/tex]
[tex]\[ 111(3.469) \approx 385.139 \][/tex]
[tex]\[ y \approx -192.544 + 385.139 + 50 \approx 242.515 \][/tex]
Thus, the vertex is at approximately [tex]\( (3.469, 242.516) \)[/tex].
Summary of Results:
- The roots of the equation [tex]\( y = -16x^2 + 111x + 50 \)[/tex] are approximately [tex]\( x \approx -0.424 \)[/tex] and [tex]\( x \approx 7.362 \)[/tex].
- The vertex of the parabola is approximately at [tex]\( (x, y) = (3.469, 242.516) \)[/tex].
Step 1: Determine the Roots
The roots of the quadratic equation can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In our equation, [tex]\( a = -16 \)[/tex], [tex]\( b = 111 \)[/tex], and [tex]\( c = 50 \)[/tex]. Plugging these values into the quadratic formula, we get:
[tex]\[ x = \frac{-111 \pm \sqrt{{111}^2 - 4(-16)(50)}}{2(-16)} \][/tex]
Calculating inside the square root first:
[tex]\[ b^2 - 4ac = 111^2 - 4(-16)(50) = 12321 + 3200 = 15521 \][/tex]
Then, the square root of 15521 is approximately:
[tex]\[ \sqrt{15521} \approx 124.599 \][/tex]
Now we compute the two potential solutions for [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{-111 + 124.599}{-32} \approx \frac{13.599}{-32} \approx -0.424 \][/tex]
[tex]\[ x_2 = \frac{-111 - 124.599}{-32} \approx \frac{-235.599}{-32} \approx 7.362 \][/tex]
So, the roots of the equation are approximately [tex]\( x_1 \approx -0.424 \)[/tex] and [tex]\( x_2 \approx 7.362 \)[/tex].
Step 2: Determine the Vertex
For a parabola expressed in the form [tex]\( y = ax^2 + bx + c \)[/tex], the x-coordinate of the vertex can be found using:
[tex]\[ x = \frac{-b}{2a} \][/tex]
Substituting the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[ x = \frac{-111}{2(-16)} = \frac{-111}{-32} \approx 3.469 \][/tex]
To find the y-coordinate of the vertex, plug [tex]\( x = 3.469 \)[/tex] back into the original equation:
[tex]\[ y = -16(3.469)^2 + 111(3.469) + 50 \][/tex]
Calculating each component:
[tex]\[ -16(3.469)^2 \approx -16(12.034) \approx -192.544 \][/tex]
[tex]\[ 111(3.469) \approx 385.139 \][/tex]
[tex]\[ y \approx -192.544 + 385.139 + 50 \approx 242.515 \][/tex]
Thus, the vertex is at approximately [tex]\( (3.469, 242.516) \)[/tex].
Summary of Results:
- The roots of the equation [tex]\( y = -16x^2 + 111x + 50 \)[/tex] are approximately [tex]\( x \approx -0.424 \)[/tex] and [tex]\( x \approx 7.362 \)[/tex].
- The vertex of the parabola is approximately at [tex]\( (x, y) = (3.469, 242.516) \)[/tex].