Graph the equation.
$y=x^2-4 x+5$
Answer (2)
The problem requires graphing the quadratic equation y = x 2 − 4 x + 5 . The vertex form is found by completing the square: y = ( x − 2 ) 2 + 1 , so the vertex is ( 2 , 1 ) . The y-intercept is found by setting x = 0 , resulting in y = 5 . There are no x-intercepts since the discriminant is negative. The graph is a parabola opening upwards with vertex at ( 2 , 1 ) and passing through ( 0 , 5 ) and ( 4 , 5 ) . The final answer is the graph of this parabola.
Explanation
Understanding the Equation We are given the equation y = x 2 − 4 x + 5 and asked to graph it. This is a quadratic equation, which represents a parabola. To graph it, we'll first find the vertex, y-intercept, and any x-intercepts.
Finding the Vertex To find the vertex, we can complete the square to rewrite the equation in vertex form, y = a ( x − h ) 2 + k , where ( h , k ) is the vertex. We have
y = x 2 − 4 x + 5
y = ( x 2 − 4 x + 4 ) + 5 − 4
y = ( x − 2 ) 2 + 1
So, the vertex is ( 2 , 1 ) .
Determining the Axis of Symmetry The axis of symmetry is the vertical line that passes through the vertex. Therefore, the axis of symmetry is x = 2 .
Finding the Y-Intercept To find the y-intercept, we set x = 0 in the original equation:
y = ( 0 ) 2 − 4 ( 0 ) + 5 = 5
So, the y-intercept is ( 0 , 5 ) .
Finding the X-Intercepts To find the x-intercepts, we set y = 0 and solve for x :
0 = x 2 − 4 x + 5
We can use the quadratic formula to find the roots: x = 2 a − b ± b 2 − 4 a c , where a = 1 , b = − 4 , and c = 5 .
x = 2 ( 1 ) 4 ± ( − 4 ) 2 − 4 ( 1 ) ( 5 ) = 2 4 ± 16 − 20 = 2 4 ± − 4
Since the discriminant is negative, there are no real x-intercepts.
Plotting Points Now we plot the vertex ( 2 , 1 ) and the y-intercept ( 0 , 5 ) . Since the axis of symmetry is x = 2 , we can find a point symmetric to the y-intercept. The y-intercept is 2 units to the left of the axis of symmetry, so there is a point 2 units to the right of the axis of symmetry with the same y-value. This point is ( 4 , 5 ) .
Finding Additional Points We can plot a few more points to get a better idea of the shape of the parabola. For example, when x = 1 , y = ( 1 ) 2 − 4 ( 1 ) + 5 = 1 − 4 + 5 = 2 . So, the point ( 1 , 2 ) is on the parabola. Due to symmetry, the point ( 3 , 2 ) is also on the parabola.
Sketching the Parabola Finally, we sketch the parabola through the plotted points. The parabola opens upwards since the coefficient of x 2 is positive.
Examples
Understanding parabolas is crucial in various real-world applications. For instance, the trajectory of a projectile, like a ball thrown in the air, follows a parabolic path. By knowing the equation of the parabola, we can determine the maximum height the ball reaches and how far it travels. Similarly, satellite dishes and reflecting telescopes use parabolic reflectors to focus signals or light to a single point, optimizing their performance.
To graph the equation y = x 2 − 4 x + 5 , identify the vertex at ( 2 , 1 ) and the y-intercept at ( 0 , 5 ) . The parabola opens upwards with no x-intercepts due to a negative discriminant. Plot these points and sketch the U-shaped curve accordingly.
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