Vertex Form of a Quadratic Equation

Key Questions

  • Answer:

    " "
    Please read the explanation.

    Explanation:

    " "
    Quadratic Equations in Vertex Form have a general form:

    color(red)(y=f(x)=a(x-h)^2+k, where

    color(red)((h,k) is the color(blue)("Vertex"

    Let us consider a quadratic equation in Vertex Form:

    color(blue)(y=f(x)=(x-3)^2+8, where

    color(green)(a=1; h=3; k=8

    Hence, color(blue)("Vertex "= (3, 8)

    To find the y-intercept, set color(red)(x=0

    y=(0-3)^2+8

    y=9+8

    y= 17

    Hence, the y-intercept: color(blue)((0, 17)

    We can use a table of values to draw the graph:

    enter image source here

    Use the table with two columns color(red)(x and y to draw the graph as shown below:

    enter image source here

    The Parent Graph of color(red)(y=x^2 can also be seen for comparison, to better understand transformation.

    Also note that,

    **Axis of Symmetry is ** color(red)(x=h

    rArr x= 3

    We can verify this from the graph below:

    enter image source here

    Hope it helps.

  • Since the equation is:

    y=x^2+bx+c

    the vertex is V(-b/(2a),-Delta/(4a)),

    or, found the x_v=-b/(2a) you can substitue it in the equation of the parabola at the place of x, finding the y_v.

  • Vertex Form

    y=a(x-h)^2+k,

    where (h,k) is the vertex.


    I hope that this was helpful.

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