Find the equation, in vertex form, of the quadratic equation with zeros -1 and 7 and through the point (5,6).
please help! i'm struggling with this
The zeroes tell you that
y = a(x+1)(x-7)
the point tells you that
a(5+1)(5-7) = 6
so a = -1/2
y = -1/2 (x^2-6x-7) = -1/2 (x^2-6x+9-16)
y = -1/2 (x-3)^2 + 8
To find the equation in vertex form of a quadratic equation, we need to use the zeros (also known as roots or x-intercepts) and a point on the graph of the quadratic.
Let's start by finding the values of a, b, and c from the given zeros (x1 = -1 and x2 = 7).
The equation with zeros x1 and x2 can be written as:
(x - x1)(x - x2) = 0
Substituting the given values, we have:
(x - (-1))(x - 7) = 0
(x + 1)(x - 7) = 0
Expanding this equation:
x^2 - 7x + x - 7 = 0
x^2 - 6x - 7 = 0
This equation gives us the standard form of the quadratic equation.
Now let's use the given point (5, 6) to find the value of a in the vertex form equation (y = a(x - h)^2 + k).
Substituting the point (5, 6) into the equation, we have:
6 = a(5 - h)^2 + k
The x-coordinate of the vertex is the average of the zeros:
h = (x1 + x2) / 2
h = (-1 + 7) / 2
h = 6 / 2
h = 3
Now substitute h = 3 into the equation:
6 = a(5 - 3)^2 + k
6 = 4a + k
We now have a system of equations:
x^2 - 6x - 7 = 0 (equation 1)
6 = 4a + k (equation 2)
To solve this system, we can use substitution or elimination. Let's use substitution.
Rearrange equation 2 to solve for k:
k = 6 - 4a
Substitute this into equation 1:
x^2 - 6x - 7 = 0
4a + (6 - 4a) = 0
4a + 6 - 4a = 0
6 = 0
Since this gives us 6 = 0, there is no solution for a.
Therefore, there is no equation in vertex form that satisfies both the zeros and the given point.
To find the equation of a quadratic function in vertex form, we need to know the zeros of the function and a point it passes through. The vertex form of a quadratic equation is given by:
y = a(x - h)^2 + k
Where (h, k) represents the vertex of the parabola.
In this case, we know the zeros are -1 and 7. Since the parabola crosses the x-axis at these points, we can write:
(x - (-1))(x - 7) = 0
Expanding this equation, we get:
(x + 1)(x - 7) = 0
Now, let's solve this equation to find the value of x:
(x + 1)(x - 7) = 0
Setting each factor equal to zero:
x + 1 = 0 or x - 7 = 0
Solving for x:
x = -1 or x = 7
So, the zeros of the quadratic equation are x = -1 and x = 7.
Now, we have the point (5,6) that the parabola passes through. We can substitute these values of x and y into the vertex form equation to find the value of 'a'.
6 = a(5 - h)^2 + k
To find 'a', we need to determine the values of 'h' and 'k' which represent the vertex.
Since we already have the zeros, the sum of the zeros, which is (-1 + 7), divided by 2 gives us the x-coordinate of the vertex. Hence:
h = (-1 + 7) / 2 = 3
Substituting this into the equation, we have:
6 = a(5 - 3)^2 + k
Simplifying further:
6 = a(2)^2 + k
6 = 4a + k
Now, we need another equation involving 'a' and 'k' to solve for both variables. We'll use one of the zeros, let's say x = -1. Plugging this value into the quadratic equation, we get:
( (-1) + 1 )((-1) - 7) = 0
Simplifying further:
(-1)(-8) = 0
8 = 0
Since this equation is not possible, we can conclude that the point x = -1 is not valid for the equation, meaning it does not lie on the parabola.
Therefore, we only have one equation left:
6 = 4a + k
Now, we need to isolate 'a' or 'k' to find their values. Let's solve for 'k' in terms of 'a':
6 = 4a + k
k = 6 - 4a
Let's substitute this value of 'k' back into the previous equation:
6 = 4a + (6 - 4a)
6 = 6
This equation always holds true, regardless of the value of 'a'. This means that there are infinitely many quadratic equations that pass through the point (5,6) with zeros at -1 and 7.
Therefore, the equation of the quadratic function in vertex form is:
y = a(x - 3)^2 + (6 - 4a)