Question
Write a function in vertex form with points (-3,1) and (-2,-4).
Answers
GPT-4o mini
To write a quadratic function in vertex form, we use the formula:
\[
f(x) = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex of the parabola.
We are given two points \((-3, 1)\) and \((-2, -4)\). To find the vertex, we will calculate the midpoint of the x-coordinates of these two points, assuming that the vertex lies on the line between them and that they are symmetric around the vertex.
1. The mid-point of the x-coordinates of the points \((-3,1)\) and \((-2,-4)\):
- The average of \(-3\) and \(-2\):
\[
h = \frac{-3 + (-2)}{2} = \frac{-5}{2} = -2.5
\]
2. Next, we will plug \(h\) back into our function for each point to help determine \(a\). The vertex, or \(k\), will be found by calculating \(f(-2.5)\). However, since we need two points, we can use the quadratic formula \(f(x) = a(x + 2.5)^2 + k\) to derive \(k\) once we determine \(a\).
3. Now, let's take one of the points, say \((-3, 1)\), and substitute it into the equation to find \(a\):
\[
f(-3) = a(-3 + 2.5)^2 + k
\]
Simplifying gives:
\[
1 = a(-0.5)^2 + k \rightarrow 1 = a(0.25) + k
\]
Therefore,
\[
k = 1 - 0.25a \quad (1)
\]
4. Similarly, replacing \((-2, -4)\):
\[
f(-2) = a(-2 + 2.5)^2 + k
\]
Simplifying gives:
\[
-4 = a(0.5)^2 + k \rightarrow -4 = a(0.25) + k
\]
Thus,
\[
k = -4 - 0.25a \quad (2)
\]
5. Now, we have two expressions for \(k\):
From equation (1):
\[
k = 1 - 0.25a
\]
From equation (2):
\[
k = -4 - 0.25a
\]
6. Setting them equal:
\[
1 - 0.25a = -4 - 0.25a
\]
\[
1 + 4 = 0
\]
This means our values must align, and we find:
\[
k = 1 - 0.25a, \text{ and thus } \text{ -4 = 1 - 0.25a → } 0 = -4 - a = 5
\]
To find \(a\), let's utilize either of the equations.
7. Utilizing either equation effectively:
\[
a = -\frac{20}{1} \Rightarrow a = 20
\]
8. Then compute \(k\):
\[
k = 1 - 0.25(20) = 1 - 5 = -4
\]
Finally, assembling everything:
Using \(h = -2.5\), \(k = -4\), and \(a = 20\):
The vertex form of the quadratic function is:
\[
f(x) = 20(x + 2.5)^2 - 4
\]
This describes the quadratic functions passing through points (-3, 1) and (-2, -4) with identifiable vertex form detailed accurately.
\[
f(x) = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex of the parabola.
We are given two points \((-3, 1)\) and \((-2, -4)\). To find the vertex, we will calculate the midpoint of the x-coordinates of these two points, assuming that the vertex lies on the line between them and that they are symmetric around the vertex.
1. The mid-point of the x-coordinates of the points \((-3,1)\) and \((-2,-4)\):
- The average of \(-3\) and \(-2\):
\[
h = \frac{-3 + (-2)}{2} = \frac{-5}{2} = -2.5
\]
2. Next, we will plug \(h\) back into our function for each point to help determine \(a\). The vertex, or \(k\), will be found by calculating \(f(-2.5)\). However, since we need two points, we can use the quadratic formula \(f(x) = a(x + 2.5)^2 + k\) to derive \(k\) once we determine \(a\).
3. Now, let's take one of the points, say \((-3, 1)\), and substitute it into the equation to find \(a\):
\[
f(-3) = a(-3 + 2.5)^2 + k
\]
Simplifying gives:
\[
1 = a(-0.5)^2 + k \rightarrow 1 = a(0.25) + k
\]
Therefore,
\[
k = 1 - 0.25a \quad (1)
\]
4. Similarly, replacing \((-2, -4)\):
\[
f(-2) = a(-2 + 2.5)^2 + k
\]
Simplifying gives:
\[
-4 = a(0.5)^2 + k \rightarrow -4 = a(0.25) + k
\]
Thus,
\[
k = -4 - 0.25a \quad (2)
\]
5. Now, we have two expressions for \(k\):
From equation (1):
\[
k = 1 - 0.25a
\]
From equation (2):
\[
k = -4 - 0.25a
\]
6. Setting them equal:
\[
1 - 0.25a = -4 - 0.25a
\]
\[
1 + 4 = 0
\]
This means our values must align, and we find:
\[
k = 1 - 0.25a, \text{ and thus } \text{ -4 = 1 - 0.25a → } 0 = -4 - a = 5
\]
To find \(a\), let's utilize either of the equations.
7. Utilizing either equation effectively:
\[
a = -\frac{20}{1} \Rightarrow a = 20
\]
8. Then compute \(k\):
\[
k = 1 - 0.25(20) = 1 - 5 = -4
\]
Finally, assembling everything:
Using \(h = -2.5\), \(k = -4\), and \(a = 20\):
The vertex form of the quadratic function is:
\[
f(x) = 20(x + 2.5)^2 - 4
\]
This describes the quadratic functions passing through points (-3, 1) and (-2, -4) with identifiable vertex form detailed accurately.