a Start by identifying a, b, and c in the given equation.
B
b Choose five points for the table. Include the x-coordinate of the vertex obtained in Part A, and use points from either side of the vertex for the remainder.
C
c Plot the points obtained in Part B and connect them with a smooth curve.
A
a y-intercept: - 52 Axis of Symmetry: x=- 43 x-coordinate of the vertex: - 43
B
bExample Table:
x
f(x)= 32x^2+4x- 52
- 4
112
- 83
- 52
- 43
- 316
0
- 52
43
112
C
cGraph:
Practice makes perfect
a Consider the general expression of a quadratic function, y=ax^2+bx+c, where a ≠0. Let's note three things we can learn from this equation.
The x-coordinate of the vertex is - b2a.We will start by identifying the values of a, b, and c.
f(x)=3/2x^2+4x-5/2 ⇔ f(x)=3/2x^2+4x+(- 5/2)
We can see that a = 32, b = 4, and c = - 52. Since the y-intercept is given by the value of c, we know that the y-intercept is - 52. Let's now substitute a= 32 and b=4 into - b2a to find the axis of symmetry and the x-coordinate of the vertex.
The equation of the axis of symmetry is x=- 43, and the x-coordinate of the vertex is - 43.
b Now, let's make a table of values using five points. We want the center point to be the vertex and the remaining points to be symmetric on either side of it. We know that the points will be symmetric if the x-coordinates are equidistant from the axis of symmetry.
x
32x^2+4x- 52
f(x)= 32x^2+4x- 52
- 4
32( - 4)^2+4( - 4)- 52
112
- 83
32( - 83)^2+4( - 83)- 52
- 52
- 43
32( - 43)^2+4( - 43)- 52
- 316
0
32( 0)^2+4( 0)- 52
- 52
43
32( 43)^2+4( 43)- 52
112
c Finally, we will graph the function by plotting the points from the table. Because the graph of a quadratic function is a parabola, we will connect them with a smooth curve.
