Graph the equation.
$y=4 x^2+8 x+7$
Answer (2)
Rewrite the equation in vertex form by completing the square: y = 4 ( x + 1 ) 2 + 3 .
Identify the vertex as ( − 1 , 3 ) and the axis of symmetry as x = − 1 .
Find the y-intercept by setting x = 0 : ( 0 , 7 ) .
Determine that there are no x-intercepts since the discriminant is negative. The final graph is a parabola with the calculated features. y = 4 x 2 + 8 x + 7
Explanation
Analyze the equation We are given the equation y = 4 x 2 + 8 x + 7 and asked to graph it. This is a quadratic equation, which represents a parabola. To graph it, we will first rewrite it in vertex form.
Factor out the leading coefficient To rewrite the equation in vertex form, we complete the square. First, factor out the coefficient of the x 2 term from the first two terms: y = 4 ( x 2 + 2 x ) + 7 .
Complete the square Now, complete the square inside the parentheses. Take half of the coefficient of the x term (which is 2), square it (which is 1), and add and subtract it inside the parentheses: y = 4 ( x 2 + 2 x + 1 − 1 ) + 7 .
Rewrite as a squared term Rewrite the expression inside the parentheses as a squared term: y = 4 (( x + 1 ) 2 − 1 ) + 7 .
Distribute Distribute the 4: y = 4 ( x + 1 ) 2 − 4 + 7 .
Vertex form Simplify: y = 4 ( x + 1 ) 2 + 3 . This is the vertex form of the equation, y = a ( x − h ) 2 + k , where the vertex is ( h , k ) .
Identify the vertex and axis of symmetry From the vertex form, we can identify the vertex as ( − 1 , 3 ) . The axis of symmetry is the vertical line x = − 1 .
Find the y-intercept To find the y-intercept, set x = 0 in the original equation: y = 4 ( 0 ) 2 + 8 ( 0 ) + 7 = 7 . So the y-intercept is ( 0 , 7 ) .
Find the x-intercepts To find the x-intercepts, set y = 0 and solve for x : 0 = 4 x 2 + 8 x + 7 . We can use the quadratic formula to find the roots: x = 2 a − b ± b 2 − 4 a c . In this case, a = 4 , b = 8 , and c = 7 . The discriminant is b 2 − 4 a c = 8 2 − 4 ( 4 ) ( 7 ) = 64 − 112 = − 48 . Since the discriminant is negative, there are no real x-intercepts.
Plot points and use symmetry Now we can plot the vertex ( − 1 , 3 ) and the y-intercept ( 0 , 7 ) . Since the parabola is symmetric about the line x = − 1 , we can find another point on the parabola by reflecting the y-intercept across the axis of symmetry. The y-intercept is 1 unit to the right of the axis of symmetry, so we can find a point 1 unit to the left of the axis of symmetry, which is x = − 2 . When x = − 2 , y = 4 ( − 2 ) 2 + 8 ( − 2 ) + 7 = 16 − 16 + 7 = 7 . So the point ( − 2 , 7 ) is also on the parabola.
Sketch the parabola Finally, we sketch the parabola through the plotted points. The parabola opens upwards since the coefficient of the x 2 term is positive.
Final Answer The graph of the equation y = 4 x 2 + 8 x + 7 is a parabola with vertex at ( − 1 , 3 ) , axis of symmetry x = − 1 , y-intercept at ( 0 , 7 ) , and no x-intercepts.
Examples
Understanding quadratic equations and their graphs is crucial in various fields. For instance, engineers use parabolas to design suspension bridges and arches, ensuring structural stability and optimal load distribution. Similarly, in physics, the trajectory of a projectile, like a ball thrown in the air, follows a parabolic path, allowing us to predict its range and height. By analyzing the equation y = 4 x 2 + 8 x + 7 , we can determine the vertex, intercepts, and axis of symmetry, which are essential for understanding the behavior and properties of the parabola, enabling us to solve real-world problems related to optimization and design.
The equation y = 4 x 2 + 8 x + 7 can be rewritten in vertex form as y = 4 ( x + 1 ) 2 + 3 , with a vertex at ( − 1 , 3 ) and a y-intercept at ( 0 , 7 ) . There are no real x-intercepts since the discriminant is negative. Thus, the graph is a parabola that opens upwards, centered on the line x = − 1 .
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