opens downward equation when the downward graph's points are on the left is -5, on the right is 5, and on top is 5

Answer 1

To find the equation of a downward-opening parabola, we can start with the general form of a quadratic function:

\[ y = a(x - h)^2 + k \]

Here, \((h, k)\) is the vertex of the parabola, and \(a\) determines the direction (positive for upwards, negative for downwards) and the width of the parabola.

Given that the vertex (the top point of the parabola) is at \((0, 5)\), we have \(h = 0\) and \(k = 5\).

The parabola opens downward, so \(a\) will be negative.

The points on the left and right are \((-5, y)\) and \((5, y)\). To find the value of \(y\) at these points, we can use the vertex form of the equation we had earlier, knowing that at \(x = -5\) and \(x = 5\), the value of \(y\) should be equal (since it's symmetric around the vertex \(x = 0\)).

Let's find \(y\):

1. Substitute \(x = 5\) into the vertex form: \[ y = a(5 - 0)^2 + 5 = a(25) + 5 \] \[ y = 25a + 5 \]

2. Substitute \(x = -5\) into the vertex form: \[ y = a(-5 - 0)^2 + 5 = a(25) + 5 \] \[ y = 25a + 5 \]

So the \(y\) values at both \((-5, y)\) and \((5, y)\) equal \(25a + 5\).

Now, since we don’t have specific information about \(y\) at those points, we can leave the equation in terms of \(a\):

If we want to specify \(y\) at \((-5)\) and \((5)\) based on some specific values of \(a\), we need \(a\) to determine the height of the parabola at those points.

As an example, let's assume \(a = -1\) for a reasonably steep downward-opened parabola:

\[ y = -1(x - 0)^2 + 5 \] \[ y = -x^2 + 5 \]

Now substituting \(x = 5\):

\[ y = -5^2 + 5 = -25 + 5 = -20 \]

and similarly for \(x = -5\):

\[ y = -(-5)^2 + 5 = -25 + 5 = -20 \]

Thus, the equation of your parabola could be:

\[ y = -x^2 + 5 \]

This equation describes a downward-opening parabola with its vertex at \((0, 5)\), and the points \((-5, -20)\) and \((5, -20)\). Adjust \(a\) based on the width or steepness you desire.

Answer 2

thats good but I want the lines a little wider and closer to the -5 and 5

Answer 3

To make the parabola wider, you can choose a value for \(a\) that is closer to zero but still negative.

For example, if we choose \(a = -\frac{1}{4}\), the vertex form of the equation will be:

\[ y = -\frac{1}{4}(x - 0)^2 + 5 \]

This simplifies to:

\[ y = -\frac{1}{4}x^2 + 5 \]

Now let's find the \(y\) values at \(x = -5\) and \(x = 5\):

1. Substitute \(x = 5\):

   \[ y = -\frac{1}{4}(5)^2 + 5 = -\frac{1}{4}(25) + 5 = -\frac{25}{4} + 5 = -\frac{25}{4} + \frac{20}{4} = -\frac{5}{4} \]

2. Substitute \(x = -5\):

   \[ y = -\frac{1}{4}(-5)^2 + 5 = -\frac{1}{4}(25) + 5 = -\frac{25}{4} + 5 = -\frac{25}{4} + \frac{20}{4} = -\frac{5}{4} \]

So, with \(a = -\frac{1}{4}\), the points on the graph at \(x = -5\) and \(x = 5\) are \((-5, -\frac{5}{4})\) and \((5, -\frac{5}{4})\).

The resulting equation is:

\[ \boxed{y = -\frac{1}{4}x^2 + 5} \]

This parabola will be wider and will have its endpoints closer to the points \((-5, -\frac{5}{4})\) and \((5, -\frac{5}{4})\). Adjust \(a\) further towards zero if you want the parabola to be even wider.