Question
A coordinate plane's x-axis ranges from 0 to 8 and its y-axis ranges from negative 4 to 4, both by 1-unit increments. A concave down parabola is drawn passing through the 1st and 4th quadrants.
Use the graph to write the equation in vertex form of the quadratic function.
(1 point)
y=−(x−)2+
Use the graph to write the equation in vertex form of the quadratic function.
(1 point)
y=−(x−)2+
Answers
GPT-4o mini
To write the equation of a concave down parabola in vertex form, we can use the formula:
\[ y = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola and \(a\) is a negative constant (since the parabola opens downwards).
For the parabola passing through the 1st and 4th quadrants on the provided coordinate plane, you will need to determine the vertex location (h, k) and the value of \(a\).
Assuming a possible vertex point where the parabola changes direction, let’s say the vertex of the parabola is at \( (h, k) = (4, 2) \) (which is a sample choice; you need to confirm this based on the graph provided). Since the parabola opens downwards, \(a\) would likely be negative. For a typical vertex form of a downward-opening parabola, \(a\) is chosen based on some specific points it passes through.
To complete the equation, insert the values:
1. If \(h = 4\) and \(k = 2\), the vertex part of the equation is:
\[
y = a(x - 4)^2 + 2
\]
2. To determine \(a\), you would need to find another point that the parabola goes through. For example, if it passes through the point \((0, 0)\) (you will need to plug in the point that the parabola actually passes through on the graph):
Substitute \(x = 0\) and \(y = 0\):
\[
0 = a(0 - 4)^2 + 2
\]
\[
0 = 16a + 2
\]
\[
16a = -2 \implies a = -\frac{1}{8}
\]
This means the vertex form of your parabola would be:
\[
y = -\frac{1}{8}(x - 4)^2 + 2
\]
Hence, the specific equation with appropriate values would be:
\[
y = -\frac{1}{8}(x - 4)^2 + 2
\]
You would replace \( \_ \) with \( 4 \) and \( \_ \) with \( 2 \):
\[
y = -\frac{1}{8}(x - 4)^2 + 2
\]
Make sure to confirm the vertex and points used based on your specific graph to finalize the values of \(h, k,\) and \(a\).
\[ y = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola and \(a\) is a negative constant (since the parabola opens downwards).
For the parabola passing through the 1st and 4th quadrants on the provided coordinate plane, you will need to determine the vertex location (h, k) and the value of \(a\).
Assuming a possible vertex point where the parabola changes direction, let’s say the vertex of the parabola is at \( (h, k) = (4, 2) \) (which is a sample choice; you need to confirm this based on the graph provided). Since the parabola opens downwards, \(a\) would likely be negative. For a typical vertex form of a downward-opening parabola, \(a\) is chosen based on some specific points it passes through.
To complete the equation, insert the values:
1. If \(h = 4\) and \(k = 2\), the vertex part of the equation is:
\[
y = a(x - 4)^2 + 2
\]
2. To determine \(a\), you would need to find another point that the parabola goes through. For example, if it passes through the point \((0, 0)\) (you will need to plug in the point that the parabola actually passes through on the graph):
Substitute \(x = 0\) and \(y = 0\):
\[
0 = a(0 - 4)^2 + 2
\]
\[
0 = 16a + 2
\]
\[
16a = -2 \implies a = -\frac{1}{8}
\]
This means the vertex form of your parabola would be:
\[
y = -\frac{1}{8}(x - 4)^2 + 2
\]
Hence, the specific equation with appropriate values would be:
\[
y = -\frac{1}{8}(x - 4)^2 + 2
\]
You would replace \( \_ \) with \( 4 \) and \( \_ \) with \( 2 \):
\[
y = -\frac{1}{8}(x - 4)^2 + 2
\]
Make sure to confirm the vertex and points used based on your specific graph to finalize the values of \(h, k,\) and \(a\).