Quads: Converting Standard to Vertex form

Transforming Quadratics from Standard Form to Vertex Form

Standard Form of A Quadratic Function

 y=ax²+bx+c

 y=ax²+bx+c

• ​when |a|<1, a is a vertical compression.

• When |a|>1, a is a vertical stretch

• When a is negative, the graph is reflected across the x-axis.

• ​c is the y-intercept

Match

Match each graph with its value of a:

|a|<1

a is negative

a is negative

|a|>1

a is negative

|a|<1

|a|>1

• |a|<1

• a is negative

• a is negative

  |a|>1

• a is negative

  |a|<1

• |a|>1

Drag and Drop

Determine the y-intercept for the following functions:



$$y=x^2-5$$ ​

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$$y=-3x^2-14x+1$$ ​

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$$y=\frac{1}{2}x^2-\frac{3}{2}x-\frac{1}{2}$$ ​

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$$y=-225x^2+342x$$ ​

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Drag these tiles and drop them in the correct blank above

 Using the quadratic formula

The Quadratic formula can be broken into useful parts:

Plugging the h value into the function for x will give you the corresponding y value (aka k)

(h, k) is the vertex of the parabola

Vertex Form of A Quadratic Function

 y=a(x-h)²+k

Math Response

What is the vertex form of the function?

$$y=-3x^2-12x+7$$

Type answer here

Correct Answer: y=-3\left(x+2\right)^2+19

Deg^°

Rad

Multiple Choice

What is the vertex of the function?

$$y=-x^2+4x-9$$

• $$\left(2,\ -13\right)$$

• $$\left(-2,\ \ -21\right)$$

• $$\left(2,\ 3\right)$$

• $$\left(2,\ -5\right)$$

Multiple Choice

What is the vertex form of the equation for the function?

$$y=-x^2+4x-9$$

• $$y=-\left(x-2\right)^2-13$$

• $$y=-\left(x+2\right)^2-21$$

• $$y=-\left(x-2\right)^2+3$$

• $$y=-\left(x-2\right)^2-5$$

Finding Vertex from Standard using Completing the Square

Finding Vertex from Standard using Completing the Square

Step 1: Add/Subtract the c from both sides (if needed): Add three to both sides

Step 2: Divide by the a on both sides (if needed): Since a is 1 we can skip this step

Finding Vertex from Standard using Completing the Square

Step 4: Factor the perfect square trinomial:

Step 5: Combine like terms on the side with y:

Finding Vertex from Standard using Completing the Square

Step 6: Solve for y: subtract 4 from both sides of the equation

Finding Vertex from Standard using Completing the Square

Add/Subtract the c from both sides (if needed):

Divide by the a on both sides (if needed):

Finding Vertex from Standard using Completing the Square

Factor the trinomial and combine like terms on the side with y:

Combine like terms on the side with y:

Finding Vertex from Standard using Completing the Square

Solve for y:

Multiple Choice

To complete the square for this function, I need to add what value to both sides of the equation?

$$y-5=x^2-10x$$

• $$100x$$

• $$-5$$

• $$25x$$

• $$25$$

Multiple Choice

I am completing the square.

Given: $$y=2x^2-20x+10$$

Step 1: $$y-10=2x^2-20x$$

Step 2: $$\frac{y}{2}-5=x^2-10x$$

Step 3: $$\frac{y}{2}-5+25=x^2-10x+25$$

Step 4: $$\frac{y}{2}-5+25=\left(x-5\right)^2$$

My next step would look like:

• $$y+20=\left(x-5\right)$$

• $$\frac{y}{2}-20=\left(x+5\right)^2$$

• $$\frac{y}{2}+30=\left(x-5\right)^2$$

• $$\frac{y}{2}+20=\left(x-5\right)^2$$

Multiple Choice

The standard form of the equation is:

$$y=2x^2-20x+10$$

The vertex form of the same equation would be:

• $$y=\left(x-5\right)-40$$

• $$y=2\left(x-5\right)^2-40$$

• $$y=2\left(x-5\right)^2-60$$

• $$y=2\left(x-5\right)^2+40$$

Multiple Choice

The standard form of the equation is:

$$y=x^2+8x+14$$

The vertex form of the same equation would be:

• $$y=\left(x+4\right)^2-2$$

• $$y=\left(x+14\right)^2-8$$

• $$y=\left(x-4\right)^2+2$$

• $$y=\left(x+4\right)^2+2$$

Multiple Choice

The standard form of the equation is:

$$y=9x^2-18x+3$$

The vertex form of the same equation would be:

• $$y=9\left(x+1\right)^2-6$$

• $$y=9\left(x-1\right)^2-6$$

• $$y=9\left(x-1\right)^2+2$$

• $$y=\left(x-1\right)^2+2$$