A parabola is a symmetrical graph that models the trajectory of something being thrown. You can imagine looking at the below, we throw a ball & it arches in a relatively symmetrical fashion.

A parabola has a number of defining features. Firstly, we have the vertex, which is the maximum or minimum point on the graph, depending which way up it is (opened upwards or opened downwards).

This vertex is in the absolute centre of the graph & we have the axis of symmetry that meets that point. Effectively, we can say that all points to the left are equidistant from the axis of symmetry to their counterparts on the other side of the line.

For example, in the top example, we have the X intercepts. These are each an identical distance from the axis of symmetry (5 on each side of the line).

As with the bottom example, we don’t have to have an X or a Y intercept.

So, if we were given the equation *-x squared + 10 (x) + 1*, how would we graph it. Well first off, we would know that it was parabola. This is because parabolas are defined in standard quadratic form. The standard form of a quadratic equation is* ax squared + bx + c. *

A definition from Your Dictionary is “A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is ax² + bx + c = 0 with a, b, and c being constants, or numerical coefficients, and x is an unknown variable.”

So, the equation we have been given follows the standard form for a quadratic equation. So, let’s break it down:

f(x) = a(x) squared + b(x) + c

In this formula:

- a and b are both coefficients, which are numerical quantities placed before a variable in an algebraic expression.
- c is a constant numerical quantity

The first thing we need to do is work out what the vertex is. If we know that, we know where the axis of symmetry is & hence, can work out where other points along the line are to draw our parabola. You can see the formula to do this below. We then take the output of *-b/2a* and use it within our quadratic function to work out the second co-ordinate of the vertex.

So now we know that the co-ordinates are (5,24) of the vertex. Which means, we know where the axis of symmetry is. We can work out what the y intercept would be to give us another point, by passing zeros into our function. In this case, we get the co-ordinates (0, -1).

That gave us our second point on the parabola. To get the third, we simply take the mirror image of the y intercept. If the y intercept is minus 5 points away from the axis of symmetry, then the symmetrical point will be plus five points from the axis, in this case (10, -1).

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