Problem 95
Question
Find the axis of symmetry for each parabola whose equation is given. Use the axis of symmetry to find a second point on the parabola whose \(y\) -coordinate is the same as the given point. $$f(x)=3(x+2)^{2}-5 ; \quad(-1,-2)$$
Step-by-Step Solution
Verified Answer
The axis of symmetry for the given parabola is \(x=-2\). A second point on this parabola with the same y-coordinate as the given point (-1,-2) is (-3,-2).
1Step 1: Recognise the format of the function
The given function is in vertex form \(f(x) = a(x-h)^{2}+k\), where (h,k) is the vertex of the parabola.
2Step 2: Find the axis of symmetry
The axis of symmetry for a parabola in the standard form is \(x = h\). So for this function, the axis of symmetry is \(x=-2\).
3Step 3: Find a second point
The second point will be equally distanced from the axis of symmetry \(x=-2\). The given point (-1, -2) lies 1 unit to the right of the axis of symmetry so the second point will be 1 unit to the left, which is (-3, -2).
Key Concepts
Vertex Form of a ParabolaParabola SymmetryStandard Form of a Parabola
Vertex Form of a Parabola
The vertex form of a parabola is an essential concept in algebra that provides a direct way to identify key attributes of a parabola. Specifically, a quadratic function written in vertex form follows the equation \( f(x) = a(x-h)^2 + k \).
The parameters \( h \) and \( k \) represent the coordinates of the parabola's vertex, making it straightforward to locate this important point on a graph. The vertex is the peak point if the parabola opens downwards (\( a < 0 \)), or the lowest point if it opens upwards (\( a > 0 \)).
In the exercise provided, the quadratic function \( f(x) = 3(x+2)^2 - 5 \) is already in vertex form. Here, the vertex can be found by identifying \( h \) and \( k \) as \( -2 \) and \( -5 \), respectively. Not only does this tell us the vertex (the point \( (-2, -5) \)), but it also aids in understanding the parabola's orientation and width based on the coefficient \( a \).
The parameters \( h \) and \( k \) represent the coordinates of the parabola's vertex, making it straightforward to locate this important point on a graph. The vertex is the peak point if the parabola opens downwards (\( a < 0 \)), or the lowest point if it opens upwards (\( a > 0 \)).
In the exercise provided, the quadratic function \( f(x) = 3(x+2)^2 - 5 \) is already in vertex form. Here, the vertex can be found by identifying \( h \) and \( k \) as \( -2 \) and \( -5 \), respectively. Not only does this tell us the vertex (the point \( (-2, -5) \)), but it also aids in understanding the parabola's orientation and width based on the coefficient \( a \).
- Orientation: Since \( a = 3 \) is positive, the parabola opens upwards.
- Width: The absolute value of \( a \) affects the parabola's stretch or compression; a larger \( |a| \) results in a narrower parabola.
Parabola Symmetry
Parabolas have a symmetrical property which is fascinating both visually and algebraically. Each parabola has an axis of symmetry, a vertical line that divides it into mirror images. For any point on one side of the parabola, there is a point with the same \( y \)-coordinate on the opposite side at the same distance from the axis of symmetry.
The axis of symmetry always passes through the vertex of the parabola. In vertex form, as we saw with \( f(x) = 3(x+2)^2 - 5 \), the axis of symmetry is simply \( x = h \), which in our example is \( x = -2 \).
To utilize symmetry in solving problems, we look for pairs of points equidistant from the axis. In our exercise, given the point \( (-1, -2) \), we know it's 1 unit away from the axis of symmetry. Therefore, its symmetrical counterpart is also 1 unit away but on the opposite side, resulting in the point \( (-3, -2) \). This concept is crucial when plotting points or finding the y-intercept of a parabola.
The axis of symmetry always passes through the vertex of the parabola. In vertex form, as we saw with \( f(x) = 3(x+2)^2 - 5 \), the axis of symmetry is simply \( x = h \), which in our example is \( x = -2 \).
To utilize symmetry in solving problems, we look for pairs of points equidistant from the axis. In our exercise, given the point \( (-1, -2) \), we know it's 1 unit away from the axis of symmetry. Therefore, its symmetrical counterpart is also 1 unit away but on the opposite side, resulting in the point \( (-3, -2) \). This concept is crucial when plotting points or finding the y-intercept of a parabola.
Standard Form of a Parabola
Apart from vertex form, parabolas can also be represented by the standard form equation, which is \( f(x) = ax^2 + bx + c \).
This format is more general and is often used when a quadratic function needs to be analyzed for its roots using methods like factoring, completing the square, or applying the quadratic formula.
While the standard form doesn't directly reveal the parabola's vertex, it gives a clearer picture of how the parabola intersects the \( y \)-axis, as \( c \) is the \( y \)-intercept. Moreover, you can determine the direction in which the parabola opens by looking at the coefficient \( a \): upwards if \( a > 0 \) and downwards if \( a < 0 \).
In converting vertex form to standard form, one would expand the squared term and simplify. Doing so provides another pathway to analyze the parabola's characteristics when a problem necessitates such a form.
This format is more general and is often used when a quadratic function needs to be analyzed for its roots using methods like factoring, completing the square, or applying the quadratic formula.
While the standard form doesn't directly reveal the parabola's vertex, it gives a clearer picture of how the parabola intersects the \( y \)-axis, as \( c \) is the \( y \)-intercept. Moreover, you can determine the direction in which the parabola opens by looking at the coefficient \( a \): upwards if \( a > 0 \) and downwards if \( a < 0 \).
- x-intercepts: The values of \( x \) that make \( f(x) = 0 \) are the roots of the equation and can be found by solving the quadratic equation.
- Axis of symmetry: For the standard form, the axis of symmetry can be calculated using the formula \( x = -\frac{b}{2a} \).
In converting vertex form to standard form, one would expand the squared term and simplify. Doing so provides another pathway to analyze the parabola's characteristics when a problem necessitates such a form.
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