Parabola, find the vertex, axis of symmetry, x & y intercepts, draw graph

Question

For the parabola y=1/2x²+x+3 find the vertex, axis of symmetry, x- and y- intercepts, and draw the graph.

 

 

Answer 1

Let's start with the y-intercept.

 

To find the y-intercept, the point where the parabola intersects the y-axis, we want to know what y is when x = 0. So, plug x = 0 into the equation and solve for y:

 

y = 1/2(0)² + 0 + 3

y = 1/2(0) + 3

y = 0 + 3

y = 3

 

The y-intercept is (0, 3).

 

 

 

Now let's find the x-intercepts.

 

To find the x-intercepts, the points where the parabola intersects the x-axis, we want to know what x is when y = 0. So, plug y = 0 into the equation and solve for y, either by factoring or using the Quadratic Formula:

 

0 = 1/2x² + x + 3

 

x = (-1 ± √((1)² - (4)(1/2)(3))) / (2(1/2))

x = (-1 ± √(1 - 6)) / 1

x = -1 ± √(-5)

x = -1 ± √(5)i

x = -1 + √(5)i    OR    -1 - √(5)i

 

Both of these roots are complex, since they involve i = √(-1). So, there are no x-intercepts.

 

 

 

Now, let's find the axis of symmetry. You can either complete the square and change the equation into vertex form, or you can use the axis of symmetry equation:

 

axis of symmetry is x = -b/(2a)

 

For your equation, b = 1 and a = 1/2, so -b/(2a) = -1/(2(1/2)) = -1/1 = -1. So, the axis of symmetry is x = -1.

 

 

 

Finally, let's find the vertex.

 

The vertex is the point on the parabola that lies on the axis of symmetry. So we can plug in x = -1 and solve for y to find the y-coordinate of the vertex:

 

y = 1/2(-1)² + (-1) + 3

y = 1/2(1) - 1 + 3

y = 1/2 - 1 + 3

y = -1/2 + 3

y = 2¹/₂

 

So, the vertex is (-1, 2¹/₂).

 

 

 

Using all that information, you can graph the points (0, 3) and (-1, 2¹/₂). The parabola will open upwards since a is positive (1/2), and it will not cross the x-axis since there are no x-intercepts. Sketch the rest of the parabola, knowing that it will be symmetric across the vertical line x = -1.