Graph the function.
$f(x)=\frac{1}{3} x^2-2 x+8$
Answer (2)
Find the vertex of the parabola using x v = − 2 a b and y v = f ( x v ) , which gives the vertex ( 3 , 5 ) .
Determine the y-intercept by setting x = 0 , resulting in the point ( 0 , 8 ) .
Check for x-intercepts using the discriminant b 2 − 4 a c . Since the discriminant is negative, there are no real x-intercepts.
Sketch the parabola using the vertex and y-intercept, noting that it opens upwards and is symmetric. The final graph is a parabola with vertex at ( 3 , 5 ) and y-intercept at ( 0 , 8 ) .
Explanation
Understanding the Problem We are asked to graph the quadratic function f ( x ) = 3 1 x 2 − 2 x + 8 . To do this, we need to find the vertex, y-intercept, and x-intercepts (if any) of the parabola.
Finding the Vertex - x-coordinate The x-coordinate of the vertex is given by x v = − 2 a b , where a = 3 1 and b = − 2 . Therefore, x v = − 2 ( 3 1 ) − 2 = 3 2 2 = 2 × 2 3 = 3.
Finding the Vertex - y-coordinate Now, we calculate the y-coordinate of the vertex by substituting x v = 3 into the function: y v = f ( 3 ) = 3 1 ( 3 ) 2 − 2 ( 3 ) + 8 = 3 1 ( 9 ) − 6 + 8 = 3 − 6 + 8 = 5. So, the vertex is at ( 3 , 5 ) .
Finding the y-intercept To find the y-intercept, we set x = 0 in the function: f ( 0 ) = 3 1 ( 0 ) 2 − 2 ( 0 ) + 8 = 0 − 0 + 8 = 8. Thus, the y-intercept is at ( 0 , 8 ) .
Finding the x-intercepts To find the x-intercepts, we set f ( x ) = 0 and solve for x using the quadratic formula: x = 2 a − b ± b 2 − 4 a c . In our case, a = 3 1 , b = − 2 , and c = 8 . The discriminant is: b 2 − 4 a c = ( − 2 ) 2 − 4 ( 3 1 ) ( 8 ) = 4 − 3 32 = 3 12 − 32 = − 3 20 . Since the discriminant is negative, there are no real x-intercepts.
Finding Additional Points Now we plot the vertex ( 3 , 5 ) and the y-intercept ( 0 , 8 ) . Since the parabola is symmetric with respect to the vertical line through the vertex ( x = 3 ), we can find another point on the parabola by reflecting the y-intercept across this line. The y-intercept is 3 units to the left of the vertex, so the reflected point will be 3 units to the right of the vertex, which is at x = 3 + 3 = 6 . The y-coordinate of this point will be the same as the y-intercept, which is 8. So, the point ( 6 , 8 ) is also on the parabola.
Sketching the Parabola Finally, we sketch the parabola through the plotted points. The parabola opens upwards, has a vertex at ( 3 , 5 ) , a y-intercept at ( 0 , 8 ) , and passes through the point ( 6 , 8 ) . There are no x-intercepts.
Examples
Understanding quadratic functions is crucial in various real-world applications. For instance, engineers use parabolas to design suspension bridges and antennas. The path of a projectile, like a ball thrown in the air, also follows a parabolic trajectory. By analyzing the vertex and intercepts of a quadratic function, we can determine the maximum height and range of the projectile, which is essential in sports and military applications. Furthermore, quadratic functions are used in economics to model cost and revenue curves, helping businesses optimize their production and pricing strategies.
To graph the function f ( x ) = 3 1 x 2 − 2 x + 8 , we find the vertex at ( 3 , 5 ) and the y-intercept at ( 0 , 8 ) . There are no real x-intercepts because the discriminant is negative. The parabola opens upward and is symmetric about the line x = 3 .
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