I was wondering if someone would be kind enough to walk me through how to do this problem:
Describe when the parabola y=-3x(power of 2) + 15x + 42 crosses the x-axis. Confirm your answer by graphing the equation.
Thank you for the help guys
I was wondering if someone would be kind enough to walk me through how to do this problem:
Describe when the parabola y=-3x(power of 2) + 15x + 42 crosses the x-axis. Confirm your answer by graphing the equation.
Thank you for the help guys
Note that when a point lies on the x-axis, it has a y-coordinate of 0. So in order to find your roots, we set your quadratic equation equal to 0, i.e.:
$\displaystyle 0 = -3x^2 + 15x + 42$
First, divide -3 from both sides since we're going to have to factor and it's much easier to deal with no coefficients in front of the $\displaystyle x^{2}$:
$\displaystyle 0 = x^{2} - 5x - 14$
Now factor the quadratic to get the form: $\displaystyle 0 = (x - a)(x-b)$ and solve for x by setting $\displaystyle x - a = 0$ and $\displaystyle x - b = 0$.
This works because when you multiply 2 numbers to get 0, either one or the other or both are equal to 0. So, setting both terms (x-a) and (x-b) equal to 0 will lead you to your x-values that correspond to the y-intercept.
Here's a smart way to think about it. You have your two x-intercepts right? This is enough information to find the x-value of your vertex since the quadratic (parabola) is symmetric about it. Your x-value for the vertex should be dead centre between your two x-intercepts.
Then, plug this x-value back into your original quadratic to find its value. Now, you have your vertex and x-intercepts which is enough to graph it. (You may also want to find the y-intercepts by simply setting x = 0 in your original quadratic).
Edit: Too slow, but here it is anyway.
Plot the x-intercepts found after you factor the quadratic.
Plot the y-intercept. In $\displaystyle y=ax^2+bx+c$, the y-intercept is "c".
Convert $\displaystyle y=ax^2+bx+c$ to vertex form and find the turnaround point. The parabola opens downward since the coefficient of $\displaystyle x^2 $is negative.
If you need more points to plot, set up a table using
x and f(x). Assign arbitrary values for x and solve for f(x).