in this lesson you will learn how to graph quadratic functions in standard form the first step is to determine whether the parabola opens upward or downward for a quadratic function in standard form if a is greater than zero the parabola opens upward if a is less than zero the parabola opens downward for our function a is 1 which is greater than zero so the parabola opens upward next find the axis of symmetry which is a vertical line that divides the parabola into the two equal hes the formula for the axis of symmetry is X = B / 2 a for our function a is 1 B is -4 substitute these values into the formula this simplifies to two right therefore the axis of symmetry is the vertical line x = 2 next find the vertex the vertex lies on the axis of symmetry so its x coordinate is the same as the axis of symmetry to find its y-coordinate substitute its x coordinate back into the original function this simplifies to -9 therefore the vertex is the 2 comma -9 next find the Y intercept the Y intercept occurs when x equals 0 so set X to 0 and solve for y this simplifies to -5 therefore the Y intercept is the point 05 next find the X intercepts the X intercepts occur when y equal 0 so set y to zero and solve for x let's use the factoring method we need to find two numbers that multiply to give -5 and add up to -4 we know that 1 * 5 = 5 but if we make 5 Negative they multiply to -5 and also add up to -4 right therefore when we Factor this quadratic equation it becomes x + 1 * x - 5 next set each factor equal to 0 and solve for x solving the first equation we find that x = -1 solving the second equation we find that x equal 5 therefore the X intercepts are the points -1a 0 and 5 comma 0 by the way if you want to master how to find the axis of symmetry the vertex and the x and y intercepts of quadratic functions please check the links in the description there are videos for each topic with plenty of examples now that we have almost all the points we need the next step is to plot them on the graph the axis of symmetry is the vertical line x = 2 the vertex is the 2 comma 9 the Y intercept is the 0a -5 the X intercepts are the points -1a 0 and 5A 0 now we add one final point which is symmetric to the Y intercept across the axis of symmetry notice that the Y intercept is two units to the left of the axis of symmetry and remember the axis of symmetry divides the parabola into two equal halves so a point 2 units to the right of the axis of symmetry is also on the parabola this point is 4 comma -5 right finally connect these five points to plot a parabola opening upward in this example we were able to easily Factor the quadratic equation when finding the X intercepts right however this is not always the case sometimes factoring can be difficult or timeconsuming even if you use the quadratic formula the discriminant might not be a perfect square or could be negative so what should we do in such cases let's see that with this example first determine whether the parabola opens upward or downward a is -1 Which is less than zero so the parabola opens downward next find the axis of symmetry using x = b / 2 a a is NE 1 B is 6 substituting these values into the formula we find that x equal 3 right next find a Vertex its x coordinate is the same as the axis of symmetry to find its y coordinate substitute 3 for X into the original function simplifying this we find that y = 6 therefore the vertex is the 3A 6 next find the Y intercept remember the Y intercept occurs when x equals 0 so set X to 0 and solve for y this simplifies to -3 right therefore the Y intercept is the point 0a -3 by the way the Y intercept always equals the constant term now if you try to find the X intercepts you will find that the quadratic is difficult to factor even if you use the quadratic formula the discriminant is not a perfect square so you may need a calculator in such cases instead of finding the X intercepts we can use other points let's first plot the points we already have and we will come back to that the axis of symmetry is the vertical line x = 3 the vertex is the 3A 6 the Y intercept is the point 0a -3 also locate Point symmetric to the Y intercept across the axis of symmetry the Y intercept is three units to the left of the axis of symmetry right so a 3 units to the right of the axis of symmetry which is 6 comma -3 is also on the parabola now to find other points take an x value that is closer to the axis of symmetry and determine its corresponding yvalue let's take x = 1 you can also take two which is even closer to the axis of of symmetry but it is easier to work with a smaller number which is why I preferred one to find its corresponding y value substitute 1 for X into the original function simplifying this we find that y equal 2 right so we have the point 1 comma 2 plot this point on the graph now just as we located a point symmetric to the Y intercept across the axis of symmetry we also locate a point symmetric to this point across the axis of symmetry this point is two units to the left of the axis of symmetry right so a point 2 units to the right of the axis of symmetry which is 5 comma 2 is also on the parabola now connect these five points to plot a parabola opening downward notice that in both this example and the previous one the parabas cross the x-axis twice but in the next example we will see a case where the parabola touches the x-axis only once please feel free to pause the video and give it to try since a is greater than zero the parabola opens upward right next find the axis of symmetry A is 4 B is 12 substituting these values into the formula we find that x = -3 next find the vertex the x coordinate of the vertex is -3 Hales right find its y-coordinate by substituting -3 for x X into the original function simplifying this we find that y equal 0 therefore the vertex is the point -3 comma 0 notice that the y-coordinate of the vertex is 0o which means the vertex is on the x-axis remember the x intercept occurs when y equals 0 so the vertex and the x intercept are the same this happens when a quadratic has only one real solution next find the Y intercept by setting X to Zero this simplifies to 9 right therefore the Y intercept is the point 0a 9 now plot these points on the graph the axis of symmetry is the line x = -3 which is the same as - 1.5 the vertex is the point -3 comma 0 the Y intercept is the point 0a 9 also locate a point symmetric to the Y intercept across the axis of symmetry the Y intercept is 1.5 units to the right of the axis of symmetry so a point 1.5 units to the left of the axis of symmetry which is -3a 9 is also on the parabola notice that even though we know the x intercept which is the same as the vertex we have only three points but when graphing a quadratic function I recommend having at least five points to plot a decent Parabola so to add two more points take an x value closer to to the axis of symmetry and find its corresponding y value and then locate a point symmetric to it across the axis of symmetry just as we did in the previous example you can take x = -1 then find its corresponding y value by substituting -1 for X into the original function simplifying this we find that y = 1 so we have the point -1a 1 plot this point on the graph then locate a point symmetric to this point across the axis of symmetry this point is 0.5 units to the right of the axis of symmetry so a0 0.5 units to the left of the axis of symmetry which is -2a 1 is also on the parabola right by the way if you want to make your graph more accurate you can add more points like this but five points are good enough to plot a decent Parabola finally connect these points to plot a parabola opening upward now let's let's look at our final example where the parabola does not cross the x-axis at all you can determine the direction of the parabola opening the axis of symmetry the vertex and the Y intercept following the same steps we used in the previous three examples I will leave the answers and you can pause the video to work on the calculations if you have any questions please feel free to leave them in the comments below then plot these points on the graph and also locate a point symmetric to the Y intercept across the axis of symmetry which is 411 notice that both the vertex and the Y intercept are below the x-axis and we know that the parabola opens downward which means the parabola does not cross the x-axis so this function has no X intercepts this happens when the quadratic has no real solutions which occurs when its discriminant is negative another similar case is when both the vertex and the Y intercept are above the X x axis and the parabola opens upward now just as in the previous two examples you have only three points so take an x value closer to the axis of symmetry and find its corresponding y value if you take x = 1 and find its corresponding y value you will have the point 1 comma -5 then plot this point on the graph and locate a point symmetric to it across the axis of symmetry finally connect these five points to plot a parabola opening downward thanks for watching I hope you found this video helpful please give it a thumbs up and consider subscribing