find the vertex find the axis of symmetry and find the x and y intercepts then draw a graph

Question

for the parabola y=1/2x^2+x+3 find the vertex find the axis of symmetry and find the x and y intercepts then draw a graph

Answer 1

y = (1/2)x² + x + 3

 

 

We need to get this equation in vertex form.

 

y = a(x - h)² + k

 

and the vertex has coordinates (h, k).

 

Since the coefficient of the leading term is 1/2, the value of a will be 1/2.

 

y = (1/2)(x - h)² + k

 

 

Now, we determine a value of h and k.  We examine the original equation and write an equivalent.

 

y = (1/2)(x² + 2x + 1) + (5/2)

 

 

If we expand this form, we will get back the original equation.  Therefore, the vertex form is

 

y = (1/2)(x + 1)² + (5/2)

 

The vertex is  (-1, 5/2).

 

Axis of symmetry is

x = -1

 

 

To find the x-intercept and y-intercept, we can use the vertex form of the equation.

 

To find the y-intercept, set x=0.

 

y = (1/2)(x + 1)² + (5/2)

y = (1/2)(0 + 1)² + (5/2)

y = (1/2) + (5/2)

y = 3

 

The y-intercept is (0, 3).

 

 

To find the x-intercept, set y=0.

 

y = (1/2)(x + 1)² + (5/2)

0 = (1/2)(x + 1)² + (5/2)

-5/2 = (1/2)(x + 1)²

 

-5 = (x + 1)²

 

±√(-5) = x + 2

 

-2 ± √(-5) = x

 

-2 + i√(5) = x             and            -2 - i√(5) = x

 

 

We have a negative number under the square-root. Therefore, we will have complex roots and the graph will NOT intercept the x-axis.