Quadratic Equations: Standard and Vertex Form

Unlike linear equations, which graph as lines, quadratic equations graph as parabolas (which look like a lot like Us or upside down Us).

There is some important vocabulary to learn about parabolas:

  • Vertex: the highest or lowest point of the parabola.  It will be on the line of symmetry.  The formula for the vertex is $x=\dfrac{-b}{2a}$.  In each of the parabolas pictured to the right, the vertex is marked with a circle.  Note that when the parabola opens down, the vertex is at the top; when the parabola opens up, the vertex is at the bottom.
  • Roots, solutions, or x-intercepts. These terms all refer to the point or points (or, in the case of "no solution" the lack of points) at which the parabola crosses the x-axis. The roots/solutions/x-intercepts in the parabolas to the right are marked with triangles.  Note that many parabolas have both of their legs cross the x-axis, so two roots (top picture); these quadratics have "two solutions." Other just touch the x-axis at the vertex: they have "one solution." Finally, quadratics like the bottom picture that do not cross the x-axis are said to have "no solution."

There are several ways to write a quadratic equation and each form gives you an "easy" way to identify some facts about a parabola.

  • Standard form is $y=ax^2+bx+c$. 
    • Whether or not the $a$ term is positive or negative tells you if the parabola opens up (positive) or down (negative). 
    • $|a|=1$ is considered a parabola of normal width. $|a|>1$ results in narrower parabolas and $|a|<1$ results in wider parabolas.
    • Use values from standard form to fill in the values in the vertex formula.
    • In standard form, c is the y-intercept of the parabola. 
    • You can also use the values from standard form to find the discriminant ($\sqrt{b^2\pm4ac}$). A quadratic with a positive discriminant has 2 real solutions (top image); a quadratic with a negative discriminant has no real solutions (bottom image). If a quadratic's discriminant=0, there is one real solution (middle image). 
  • Vertex form is $y=a(x-h)^2+k$ (Note: $a$ is the same in both standard and vertex form.)
    • The vertex is $(h,k)$.
    • The axis of symmetry=$h$.
 

$-3x^2+5x+15$ 
 

 

 

 

$x^2+6x+9$

 

 

$x^2+4x+6$

 

 

 

 

 

 

  • To convert from standard to vertex form, you can plug in $(h,k)$, which you can derive from $-\dfrac{b}{2a}$, and $a$ into vertex form.
 
$\eqalign{y&=x^2-4x+6\\\text{vertex}=(x,y), x&=\dfrac{-b}{2a}=\dfrac{4}{2}=2\\\text{Plug x value into equation to find y of vertex}\\y&=2^2-4(2)+6\\y&=4-8+6\\y&=2\\\text{vertex}&=(2,2), h=2, k=2\\\text{Plug vertex in as (h,k)}\\y&=a(x-h)^2+k\\y&=1(x-2)^2+2\\y&=(x-2)^2+2}$

 

 

  • To convert from vertex to standard form, use $(h,k)$ and $a$ to solve for $b$ (using $-\dfrac{b}{2a}$) and then $c$.
 
 $\eqalign{y&=2(x+2)^2-3\\\text{Find vertex and plug x and a into vertex formula to solve for b}\\(h,k)&=(-2,-3), a=2\\-2&=-\dfrac{b}{2a}\\-2&=-\dfrac{b}{2(2)}\\-2&=-\dfrac{b}{4}\\-8&=-b\\b&=8\\\text{Plug a and b into standard form}\\y&=2x^2+8x+c\\\text{Plug any point (like vertex) in for x and  to solve for c}\\-3&=2(-2)^2+8(-2)+c\\-3&=-8+c\\c&=5\\y&=2x^2+8x+5}$

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