polynomials: find vertex, intercepts of f(x) = 2x^2 - 7x - 4

[bandaidgirl]

I need to find the vertex, the x- and y-intercepts, and the axis of symmetry for the following:

. . .f(x) = 2x^2 - 7x - 4

Is the vertex at the point (7,4)?

Please explain each step, if you can. Thank you!

 

[xomandi]

Since this is a positive function, the parabola opens upward, so your vertex can not be at (7, 4).

The standard form of your function is f(x) = ax^2 + bx + c. The c-value is your y-intercept (where your parabola crosses the y axis).

To find your x-intercepts, you need to set the function equal to zero and factor:

. . .(2x + 1)(x - 4)

You can check your factoring by multiplying the factors back together, and verifying that you get what you started with.

Once the quadratic is factored, set these factors equal to 0 and solve for x to find your zeros (or x-intercepts)

I don't recall at the moment how to find the vertex from the equation.

Hope this helped a bit.

 

[stapel]

To find the vertex, your book and/or class would have covered at least one method. Are you using the completing-the-square method, or the formula? Or both?

Since the axis of symmetry for a quadratic y = ax² + bx + c with vertex at (h, k) is just the line x = h, the axis is simple to write down once you've found the vertex.

Please explain each step, if you can.

Teaching lessons is beyond what we can do here. If you would like, however, we can provide links to online lessons. Please specify the topic(s) you need, if this is what you are requesting.

Thank you.

Eliz.