Home > Algebra > Vertex Form of a Quadratic Equation

How do you find the vertex and intercepts for #y= 1/2x^2#?

Answer 1

Eli Assouline

This is trivial #v(0,0)# can you see why

#h = -b/(2a)#

#k = f(h) #

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Eva Abramson

To find the vertex and intercepts for the equation ( y = \frac{1}{2}x^2 ):

Vertex: The vertex of a quadratic function in the form ( y = ax^2 + bx + c ) is at the point ((h, k)), where (h = -\frac{b}{2a}) and (k) is the value of the function at (h).

For ( y = \frac{1}{2}x^2 ), (a = \frac{1}{2}), (b = 0). Substituting into the vertex formula: ( h = -\frac{b}{2a} = -\frac{0}{2 \cdot \frac{1}{2}} = 0 ) Evaluate ( y ) at ( h = 0 ): ( y = \frac{1}{2}(0)^2 = 0 )

So, the vertex is at ((0, 0)).
Intercepts:
- x-intercept: Set (y) to zero and solve for (x). (0 = \frac{1}{2}x^2 ) (x^2 = 0 ) (x = 0 )
  
  So, the x-intercept is at ((0, 0)).
- y-intercept: Set (x) to zero and solve for (y). (y = \frac{1}{2}(0)^2 = 0 )
  
  So, the y-intercept is also at ((0, 0)).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.