Create your own quadratic equation that has an axis of symmetry of x = −1 and whose graph opens down. Then find the vertex, domain, range, and x-intercepts. Show all work to receive full credit.
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your parabola would be
y = a(x+1)^2 + c
where a would be a negative number.
So choose any values for a and c
e.g. one possible answer:
y = -2(x+1)^2 + 5
vertex: (-1,5)
domain: any real value of x
range : y ≤ 5
x-intercept: let y=0
-2(x+1)^2 +5=0
(x+1)^2 = 5/2 or 10/4
x+1 = ± √10/2
x = -1 ± √10/2
notice my little trick of making my denominator a perfect square, so it would come out "nice" after taking the square root ?
To create a quadratic equation with an axis of symmetry of x = -1 and a graph that opens down, we need to start with the general form of a quadratic equation:
y = ax^2 + bx + c
Since the axis of symmetry is x = -1, we know that the quadratic equation will be symmetric about the vertical line x = -1. This means that the x-coordinate of the vertex will be -1.
The vertex form of a quadratic equation is given by:
y = a(x - h)^2 + k
Where (h, k) represents the coordinates of the vertex. Since the x-coordinate of the vertex is -1, h = -1.
Substituting these values into the vertex form equation, we get:
y = a(x - (-1))^2 + k
y = a(x + 1)^2 + k
Now, we need to determine the value of 'a' in order to complete the equation. We know that the graph opens down, which means that the coefficient 'a' must be negative.
Let's assume a = -1 for this example. Substituting this value into the equation, we get:
y = -(x + 1)^2 + k
Now, let's find the value of 'k' by substituting the coordinates of the vertex (-1, k) into the equation:
0 = -(0 + 1)^2 + k
0 = -1 + k
k = 1
So, the quadratic equation with an axis of symmetry of x = -1 and a graph that opens down is:
y = -(x + 1)^2 + 1
Now, let's find the vertex, domain, range, and x-intercepts for this equation.
To find the vertex, we simply need to use the formula (-b/2a, f(-b/2a)), where f(x) is the equation. In our case, b = 1 and a = -1. Substituting these values, we get:
x-coordinate of vertex = -(-1) / 2(-1) = 1/2
y-coordinate of vertex = -((1/2) + 1)^2 + 1 = -9/4
So, the vertex of the quadratic equation is (1/2, -9/4).
The domain of a quadratic equation is the set of all real numbers, so the domain of this equation is (-∞, ∞).
To find the range, we need to determine the maximum or minimum point in the graph. Since the graph opens down, the vertex (-1/2, -9/4) represents the maximum point. Therefore, the range is (-∞, -9/4].
Finally, to find the x-intercepts, we set y = 0 in the equation:
0 = -(x + 1)^2 + 1
(x + 1)^2 = 1
x + 1 = ±√1
x = -1 ± 1
So, the equation has two x-intercepts: x = 0 and x = -2.
In summary, the quadratic equation with an axis of symmetry of x = -1 and a graph that opens down is:
y = -(x + 1)^2 + 1
The vertex is (1/2, -9/4), the domain is (-∞, ∞), the range is (-∞, -9/4], and the x-intercepts are x = 0 and x = -2.