Question

4. Esmeralda is graphing a polynomial function as a parabola. Before she begins graphing it, explain how to find the vertex. Make sure you include how to determine if it will be a maximum or minimum point. Use an example quadratic function to help you explain and provide its graph.

Answer 1

@austinL

Answer 2

well the easiest way if you know the roots of zeros is find the midpoint between the zeros...

this will be the line of symmetry... x = h

then substitute the value into the original equation... and get y = k

so the vertex will be (h, k)

if you have an equation in the form

\[y = ax^2 + bx + c\]

the line of symmetry is

\[x = \frac{-b}{2a}\]

then substitute it to find y...

hope it helps

Answer 3

But I don't understand how to solve one of those equations.....

Answer 4

oops forgot to say, parabolas are symmetric... and the vertex lies on the line of symmetry

Answer 5

Here's my work:

f(x) = 3x2 - x + 5
f(x) = 3(x2 – x) + 5
That's all I could do.

Answer 6

*3x^2

Answer 7

@campbell_st

Answer 8

ok... so you're attempting to put the equation into vertex from

I wouldn't worry

using the line of symmetry

\[x = \frac{-(-1)}{6}\]

substitute it to find y = 59/12


vertex (1/6, 59/12)

I just find this easier then completing the square

Answer 9

My problem is how do I solve this?
f(x) = 3x^2 - x + 5

Answer 10

well you don't the curve is positive definite... it doesn't cut the x axis..

so it has complex roots...

you can test by using the discriminant

b^2 - 4ac

(-1)^2 - 4 *3 * 5 = -59

so the equation have no real roots...