The Effects that the Coefficients of a Quadratic Equation Have on the Vertex of Its Parabola

Consider the general equation of a parabola: We know that the graph of a quadratic will give us a parabola. How will changing a,b, and c affect the vertex of the parabola?

First let's look at the standard form for the equation of a parabola: The vertex in the above form is at (h,k). We can put the general equation in terms of a,b, and c in the standard form by completing the square:  As we can see b and a will affect the x coordinate of the vertex, while a,b,and c all affect the y coordinate.

Now let's look at some graphs: First we will let b=0,c=0. From the solutions for h and k we expect the vertex to be at (0,0).  We know that with b and c equal to zero our vertex will be at (0,0), so what affect will changing a have on the parabola? Lets change a and look at the graphs: The purple graph is a=.25,the red a=.5,the blue a=.75,the green a=1 and the aqua a=1.5. So we see as a increases the parabola gets thinner. If we look at the negatives for the previous values of a we get: So negative values of a cause the parabolas to open downwards. Remember the coordinates of the vertex in terms of a,b,c where are: So let a=1 and b = 0. As we change c we expect the y coordinate of the vertex to change. As c increases the vertex will get higher. Lets look at a few parabolas: The purple parabola was obtained when c = 0, the red when c = 1 and the blue when c = - 1.

Next we will look at the cases when a=1, c=0 and b varies. The coordinates of the vertex will be: when we plot these as x,y coordinates with b varying we get: This is the locus of vertices for a=1 and c=0 with b varying. If we look at several parabolas on top of the locus of vertices we see:

red:b = 0,blue:b = 1,green:b = -1,aqua:b = 2,yellow:b=-2,lavender:b = 3,and,gray:b = -3: Now if we keep a=1 and vary c and b the coordinates of our vertex would be: This will shift our parabola up by the constant c. If we look at the previous graph with c= 1, instead of zero, we get: Our locus of vertices is now not correct. If we move it up 1 unit will it be correct?

Now for for c = -1 we would expect to shift the family of parabolas with b varying and a = 1 down one: Now what happens when we vary a? Lets look at the values of the coordinates of the vertex in terms of the coefficients of our quadratic: .

As we can see changing a will affect both the x and y-coordinates of the vertex. Lets look at the family of parabolas with b varying and c=0 with different values of a. First with a=2 we will get the following: We can compare this to the graphs when a=1,c=0: The equation of the vertex in the first graph with a=2,c=0 is and for the second graph with a=1,c=0 the equation of the vertex is . So we see that the parabola with a=2,c=0 has a vertex with coordinates that are one-half the coordinates of the vertex when a=1,c=0.

Summary

We looked at the graphs of as we changed a,b,c. We saw changing a with b,c,constant we changed the width of our parabola. As we let a go to zero our parabola increased in width. For a positive a the parabola opened upwards and and a negative a caused our parabola to open downwards. We saw that changing c caused a vertical shift upwards of the vertex by the value c and downwards the value c for negative c. When we changed b with a and c constant we found that the locus of vertices of our parabola was a parabola itself.

Further Investigation

We saw the locus of vertices of our parabola given by, ,

with a positive and b varying was a parabola that opened downwards. What affect would a negative a have on the locus of vertices parabola?

Recall the standard form of a parabola:. When we put our equation , , in standard form we got .

We see that . If y = k-p is the equation of the directrix of the standard parabola and the focus is at (h,k+p) how does a,b,c affect the directrix and focus?