The vertex (turning point) is on the axis of symmetry, so the x-coordinate is - b2 a.
Let's identify a, b, and c in the equation we would like to graph.
y=x^2+9= 1x^2+ 0x+ 9
⇓
a= 1, b= 0, c= 9
Using these values, we can find the x-coordinate of the vertex.
x-coordinate of the vertex: - 0/2( 1)=0
Let's use this value and pick numbers to the left and to the right to make a table of points on the graph of y=x^2+9.
x
y=x^2+9
Point on Graph
^2+9=9
( ,9)
1
1^2+9=10
(1,10)
2
2^2+9=13
(2,13)
-1
(-1)^2+9=10
(-1,10)
-2
(-2)^2+9=13
(-2,13)
To draw the graph, let's put these points in a coordinate system and connect them with a smooth curve.
Since this is a quadratic graph, we need to draw a parabola.
c We can follow an approach similar to Part A. This time we can use the given 2+i and 2-i to write the factor form of the quadratic equation with these two numbers as roots.
(x-(2+i))(x-(2-i))=0Like in Part A, we can expand the left hand side.
We found a quadratic equation in standard form with roots 2+i and 2-i.
x^2-4x+5=0
d We can use a method similar to Part B to draw the graph of the quadratic function y=x^2-4x+5. We start with finding the x-coordinate of the vertex using the formula - b2a.
x-coordinate of the vertex: -(-4)/2(1)=2
Next, we find the vertex and other points on the graph.
x
y=x^2-4x+5
Point on Graph
^2-4( )+5=5
( ,5)
1
1^2-4(1)+5=2
(1,2)
2
2^2-4(2)+5=1
(2,1)
3
3^2-4(3)+5=2
(3,2)
4
4^2-4(4)+5=5
(4,5)
Finally, we can plot these points and join them with a smooth parabola.
e Let's recall that x-intercepts of y=f(x) correspond to solutions of f(x)=0. We can use this observation to write a statement about quadratics.
If the graph ofy=ax^2+bx+c
has nox-intercepts,
then the quadratic equationax^2+bx+c=0
has only complex solutions.
This is confirmed by the graphs of Part B and Part D.
