We are asked to solve the given by graphing. There are three steps to solving a quadratic equation by graphing.
- Write the equation in , ax2+bx+c=0.
- Graph the related y=ax2+bx+c.
- Find the , if any.
The solutions of
ax2+bx+c=0 are the
x-intercepts of the graph of
y=ax2+bx+c. First, we have to write our equation in the standard form.
x2=3x−3⇔x2−3x+3=0
Now that we have our equation written in the standard form, let's identify the function related to the equation.
Equation: Related Function: x2−3x+3=0y=x2−3x+3
Graphing the Related Function
To draw the graph of the related function written in , we must start by identifying the values of
a, b, and
c.
y=x2−3x+3⇔y=1x2+(-3)x+3
We can see that
a=1, b=-3, and
c=3. Now, we will follow four steps to graph the function.
- Find the .
- Calculate the .
- Identify the and its across the axis of symmetry.
- Connect the points with a .
Finding the Axis of Symmetry
The axis of symmetry is a with equation
x=-2ab. Since we already know the values of
a and
b, we can substitute them into the formula.
The axis of symmetry of the parabola is the vertical line with equation
x=1.5.
Calculating the Vertex
To calculate the vertex, we need to think of
y as a function of
x, y=f(x). We can write the expression for the vertex by stating the
x- and
y-coordinates in terms of
a and
b.
Vertex: (-2ab,f(-2ab))
Note that the formula for the
x-coordinate is the same as the formula for the axis of symmetry, which is
x=1.5. Therefore, the
x-coordinate of the vertex is also
1.5. To find the
y-coordinate, we need to substitute
1.5 for
x in the given equation.
y=x2−3x+3
y=1.52−3(1.5)+3
y=2.25−3(1.5)+3
y=2.25−4.5+3
y=0.75
We found the
y-coordinate, and now we know that the vertex is
(1.5,0.75).
Identifying the y-intercept and its Reflection
The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Therefore, the point where our graph intercepts the y-axis is (0,3). Let's plot this point and its reflection across the axis of symmetry.
Connecting the Points
We can now draw the graph of the function. Since a=1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.
Finding the x-intercepts
Let's identify the x-intercepts of the graph of the related function.
We can see that the parabola does not intersect the x-axis. Therefore, the equation x2−x+3=0 has no solution. This also means that there is no solution to our original equation, x2=3x−3.