Find and plot the y-intercept and its symmetric point 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+5
⇕
y= 1x^2+ x+5
We have identified that a= 1, b= , and c=5.
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.
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, we can substitute 0 for x in the given equation.
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 intersects the y-axis is (0,5). As we see, the y-intercept is the same as the vertex, and then it will not help us draw the graph. Instead, we will find the y-coordinate of the points where x=- 2 and x=2. Let's start with x=-2.
Since a=1, which is greater than zero, we know that our parabola opens upwards.
Let's draw a smooth curve. We should not use a straight edge for this!
Extra
A Common Mistake
One common mistake when identifying the key features of a parabola algebraically is forgetting to include the negatives in the values of these constants. The standard form is addition only, so any subtraction must be treated as negative values of a, b, or c. Let's look at an example.
ax^2 + bx + c [1em]
y=3x^2-4x-2 ⇕ y=3x^2 + (-4x) + (-2)
In this case, the values of a, b, and c are 3, -4, and -2. They are not 3, 4, and 2.
cccc
a=3,& b= 4, & c= 2 & *
a=3,& b=-4,& c=-2 & âś“
