Unit #3 Quadratics

Standard Form

Learning Goals

Throughout this unit, you will learn how to:
  • Find the x-intercepts using the quadratic formula
  • using the AOS value to find the y- intercept
  • completing the square (finding the MAX or MIN value
  • solve word problems using the quadratic formula/ completing the square


This final unit is very simple. The main concepts to go over are using the quadratic formula and completing the squares. These methods help you find the values that will ultimately make it easier for you to graph your parabola. This unit connects back to Unit #1, using VERTEX FORM. Completing the square is simply converting the standard form equation to vertex form in order to find the vertex of the parabola. In order to do this, you must follow along with the steps provided below. The quadratic formula will help you find the x-intercepts of the equation which you can further use to find the axis of symmetry. Thus, you will have the x-value of your vertex. Plug the AOS value into the original standard form equation and you should find the y-intercept of your parabola. This is one way of finding the vertex besides completing the square. Within the quadratic formula, you can find the formula for the DISCRIMINANT which tells you how many solutions can be found from an equation. If your discriminant is more and 1, there will be 2 solutions. If your discriminant is 0, there will be 1 solution and if your discriminant is less and 0, there will be no solution. The discriminant formula is D= b²- 4ac.

Some key points to remember are:

  • That the value of a gives you the shape and direction of opening of the parabola
  • The value of c is y-intercept of the parabola
  • You can rewrite a quadratic relation of the form y= ax² + bx+c in the form y=a(x-h)+k by completing the square
  • The values (h,k) in vertex form represent the maximum and minimum point of the parabola
  • The vertex is a minimum point when a>0 and a maximum point when a<0.

Quadratic Formula

The quadratic formula is used when a certain quadratic cannot be factored. This formula was developed by completing the square and solving the quadratic ax² + bx + c = 0. That is also the standard form equation. The formula gives us the solutions or roots of the quadratic equation.
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This formula is modeled from the standard form equation. All you must do to find the x-intercepts, zeroes or solutions, is to substitute the a, b, and c values into the equation. Once you plug those values in, solve for x.


Find the x-intercepts, the vertex, and the equation of the axis of symmetry of the quadratic relation y= - 5x² + 8x - 3. Sketch the parabola.

*The first thing you should do I write down everything you know just by looking at the equation.*

a= -5 b= 8 c= -3

*Now substitute the values into the equations.*

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Now that you have your two x-intercepts, you can plug these into the AOS equation to find the Axis of Symmetry or in other words, your x-value for the vertex.

Once you find the AOS value, substitute that into the original standard form equation to solve for the y-intercept, as shown below.

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To recap, we have found the 2 x-intercepts, the axis of symmetry, and the vertex of this parabola. Use this information to draw a brief sketch of your parabola.
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Word Problem using the Quadratic Formula

The path of a basketball after it is thrown from a height of 1.5m above the ground is given by the equation h = -0.25d² + 2d +1.5, where h is the height , in metres, and d is the horizontal distance, in metres.

a) How far has the ball travelled horizontally, to the nearest tenth of a metre, when it lands on the ground?

*Keep in mind that there can never be a negative distance, so you must use the positive solution.*

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b) Find the horizontal distance when the basketball is at a height of 4.5m above the ground.
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The ball is at a height of 4.5m off the ground at the horizontal distances of 2m and 6m.
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The Discriminant

The discriminant can be used to determine how many solutions an equation will have. The formula for finding the discriminant is D= b²- 4ac. The discriminant can be used to determine how many solutions an equation will have. The formula for finding the discriminant is D= b²- 4ac. You must substitute the values from your original equation to solve for the discriminant. If the discriminant is more than 0 that means that there will be 2 solutions. If the discriminant is 0 then there will be 1 solution. Finally, if the discriminant is less than 0 there will be no solution (in other words, no x-intercepts).






This means that there 2 solutions.




This means there 1 solutions.





This means there are no solutions.

Completing the Square

The process of completing the square involves changing the first two terms of a quadratic relation of the form y= ax² + + c into a perfect square while maintaining the balance of the original relation.


This method is used to find the vertex form of the standard form equation. It is also really useful for finding the maximum or minimum value of an equation. To complete the squares you must first make sure that the value of a is 1. If it is not, then you must factor it out of the equation. The next step is to divide the value of b by 2 and then square it. Add this value beside b. Then you must place brackets around your equation while leaving c on the outside. Next, take the number you got from dividing b by 2 and squaring it, and subtract it (or add it depending on the situation) from c. this is complete, simply your equation.

Example: Rewrite y= x² + 8x + 5 in the form y= a(x-h)² + k.

y= x² + 8x + 5

y= (x²+ 8x) + 5


8/2= (4)² =16

y= (x²+ 8x +16 -16) +5

y= (x²+ 8x +16) -16 + 5

y= (x + 4)² - 11

By looking at the vertex form equation, you can easily find the vertex ( -4, -11) which can help you graph your parabola. This vertex tells you the MINIMUM value of the parabola.

For more examples, watch the video below.

Word Problems using Completing the Squares

The path of a ball is modelled by the equation y= -x² + 2x +3, where x is the horizontal distance, in metres, from a fence and y is the height, in metres, above the ground.

a) What is the maximum height of the ball, and at what horizontal distance does it occur?

*Do NOT forget to write your therefore statement at the end.*

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Throughout the quadratics unit, I have learned a lot of techniques that help me solve problems correctly and efficiently. We learned how to use each aspect of quadratics to construct parabolas that can be stretched, compressed, and upside down. Personally, I started off the semester with graphing being my least favourite part. However, the various methods such as "The Step Method", have helped me improve my graphing skills. It is now one of my favourite part of quadratics because all the equations we have learned provide you with ALL the necessary information needed in order to graph a parabola. All you have to do is find the x-intercepts, axis of symmetry, vertex, and optimal value and plot these onto the graph. The quadratic formula and completing the square make find essential points quick and the result is something that we see in our everyday lives, parabolas. I had not noticed it before, however, parabolas can be seen everywhere. Whether it is a rollercoaster, the Eiffel Tower, or a rainbow, parabolas are a part of our lives. Learning techniques that helped to graph parabolas were simple, however, I noticed that I made silly mistakes that could have easily been prevented. Looking back on my tests and thinking tasks, I noticed that I lost marks in thinking and application based on mistakes such as missing a negative sign or forgetting to add one. My solution to this is to take my time when solving equations and word problems, to ensure that I can successfully complete my work to the best of my abilities. I look towards improving my skills in the future. I think the Vertex Form equation was my favourite out of the 3 units because it clearly told you the vertex of the parabola. Factored form required a bit of work however, it told you the solutions/ x-intercepts right away. Finally, the standard form gives you the y-intercept which is the c value of the equation. The connections between these units can be seen through the parts of all these formulas, which help you create a parabola. Furthermore, all of the forms have the 'a' in common, which we know can only be 1 or more, never 0. Also, when completing the squares we change between the vertex and factored form, which explains how they are linked. In conclusion, all 3 units connect to each other!


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