The solutions to the equation are the x-coordinates of the points of intersection of the graph and the x-axis.
x=- 4 and x=4
Practice makes perfect
We want to solve the given quadratic equation by graphing the related function. The solutions of these equations are the x-coordinates of the points of intersection of the parabola and the x-axis. Recall that a quadratic equation can have two, one, or no solutions.
Graphing the Related Function
We want to draw the graph of a quadratic function written in standard form.
y=ax^2+bx+c
To do so, we will follow five steps.
Find and plot two other symmetric points across the axis of symmetry.
Draw a smooth curve through the three plotted points.
Let's do it!
Identify a, b, and c
We will start by identifying the values of a, b, and c.
y=x^2-16
⇕
y= 1x^2+ x+(-16)
We have identified that a= 1, b= , and c=-16.
Axis of Symmetry
The axis of symmetry is the vertical line that divides the parabola into two mirror images. Its equation follows a specific formula.
x=- b/2 a
Let's substitute our given values a= 1 and b= into this equation.
The axis of symmetry is the line x=0, thus the graph is symmetrical about the y-axis.
Calculating the Vertex
To find the vertex of the parabola, we will 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: ( - b/2a, f(- b/2a ) )
When determining the axis of symmetry, we found that - b2a=0. Therefore, the x-coordinate of the vertex is 0 and the y-coordinate is f(0). To find this value, substitute our x-coordinate for x in the given equation.
Since the graph is symmetric about the y-axis, we can find the axis of symmetry by plotting two points which are equal in distance from x=0. We will find such points by substituting - 3 and 3 into the function rule. The resulting (-3,f(- 3)) and (3,f(3)) are the coordinates of the points.
x
x^2-16
y=x^2-16
- 3
( - 3)^2-16
- 7
3
3^2-16
- 7
We found that (- 3,- 7) and (3,- 7) lie on the graph.
Graph
Since a=1, which is greater than zero, we can confirm that our parabola opens upwards.
Let's draw a smooth curve connecting the three points we have. You should not use a straight edge for this!
Determining the Real Solutions
Let's consider the graph.
This graph crosses the x-axis twice, so its related equation, x^2-16=0, has two solutions. We can see that those solutions are x=- 4 and x=4.
