Question
What is a vertex form of The graft equation y = -2 (x - blank) squared plus blank there is a point at negative 3, 1
Answers
Answered by
GPT-4o mini
The vertex form of a quadratic equation is given by the formula:
\[
y = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex of the parabola, and \(a\) is a coefficient that determines the direction and width of the parabola.
From your equation \(y = -2(x - \text{blank})^2 + \text{blank}\), we can see that the coefficient \(a = -2\).
Given that there is a point at \((-3, 1)\), we can substitute this point into the vertex form to find \(h\) and \(k\).
1. Substitute \(x = -3\) and \(y = 1\):
\[
1 = -2(-3 - h)^2 + k
\]
2. Rearranging gives us:
\[
-2(-3 - h)^2 + k = 1
\]
Now, we can express \(k\) as follows:
\[
k = 1 + 2(-3 - h)^2
\]
At this stage, we need to find \(h\) and \(k\). We have one variable in terms of another. Since we need to find a specific vertex, we can look for integer values.
To find a reasonable pair for \((h, k)\), let's first assume \(h\) is at the vertex and see if \(k\) remains within a normal range. The symmetry of quadratic equations suggests \(h\) may partially be defined by the x-coordinate of the known point.
For simplicity, let's trial an average value around -3. Trying \(h = -3\) or values around -3 would give us the corresponding \(k\):
Assuming \(h = -3\):
\[
k = 1 + 2(0)^2 = 1
\]
If \(h = -3\), then:
\[
y = -2(x + 3)^2 + 1
\]
Confirming for other points or estimating shows no valid need for adjustment, thus:
\[
y = -2(x + 3)^2 + 1
\]
Here it is correctly in vertex form.
Thus, we have for your blanks:
- The value of \(h = -3\)
- The value of \(k = 1\)
The equation in vertex form is:
\[
y = -2(x + 3)^2 + 1
\]
\[
y = a(x - h)^2 + k
\]
where \((h, k)\) is the vertex of the parabola, and \(a\) is a coefficient that determines the direction and width of the parabola.
From your equation \(y = -2(x - \text{blank})^2 + \text{blank}\), we can see that the coefficient \(a = -2\).
Given that there is a point at \((-3, 1)\), we can substitute this point into the vertex form to find \(h\) and \(k\).
1. Substitute \(x = -3\) and \(y = 1\):
\[
1 = -2(-3 - h)^2 + k
\]
2. Rearranging gives us:
\[
-2(-3 - h)^2 + k = 1
\]
Now, we can express \(k\) as follows:
\[
k = 1 + 2(-3 - h)^2
\]
At this stage, we need to find \(h\) and \(k\). We have one variable in terms of another. Since we need to find a specific vertex, we can look for integer values.
To find a reasonable pair for \((h, k)\), let's first assume \(h\) is at the vertex and see if \(k\) remains within a normal range. The symmetry of quadratic equations suggests \(h\) may partially be defined by the x-coordinate of the known point.
For simplicity, let's trial an average value around -3. Trying \(h = -3\) or values around -3 would give us the corresponding \(k\):
Assuming \(h = -3\):
\[
k = 1 + 2(0)^2 = 1
\]
If \(h = -3\), then:
\[
y = -2(x + 3)^2 + 1
\]
Confirming for other points or estimating shows no valid need for adjustment, thus:
\[
y = -2(x + 3)^2 + 1
\]
Here it is correctly in vertex form.
Thus, we have for your blanks:
- The value of \(h = -3\)
- The value of \(k = 1\)
The equation in vertex form is:
\[
y = -2(x + 3)^2 + 1
\]