The Shifted Form Of A Parabola Homework

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In the bottom-right figure, we are given a parabola that is not in standard form. The vertex is found by calculating the point at which the parabola is at its lowest (or highest) point, and the focus and directrix can be found easily. The vertex of the parabola is found at the intersection of the equation for the focus and the equation for the directrix.

We can now use the general equation we have from the vertex to the focus inorder to get our two lines. We have our down-facing parabola(x- h)2 = 4p(y - k) form and we need to isolate the x2term again. The focus is the point where the parabola intersects the x-axis, so if we set the equation equal to zero, we get the point at which the x-axis intersects the parabola: {eq}(x- h) = 0 {/eq}. This is equivalent to {eq}x = h {/eq}.

The parabola is both downward facing and downward sloped, so the graph of the function intersects the y-axis at two points. The y-intercepts are the points where the graph of the function intersects the y-axis. The two points at which the graph of the function intersects the y-axis are called the lower vertex and the higher vertex of the parabola.

The directrix is the line on which the graph of the function is parallel to. It is perpendicular to the tangent at the vertex. The vertex is the point where the tangent is at its lowest (or highest) point, and the directrix is found by finding the x-intercepts. When the foci and directrix are found in this fashion, we can find the vertex of the parabola by combining the two equations as follows:

We will first look at the two-term parabola with b positive. We nowhave an equation of the form y = a(x - h)2 + b, so our two terms x2 and (y -k)2 are both in the form (x - h)2. We will consider two cases:

when the coefficients are negative

when the coefficients are positive

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