Question

f(x)=4x^2-7x-15What are the x-intercepts of the graph of f(x)? Is the vertex going to be max or min? what are the coordinates of the vertex? what steps would I use to graph f(x)?

Answer 1

Solution

We are given the function

$f(x)=4x^2-7x-15$

To find the x - intercept, we will put f(x) = 0

$\begin{gathered} 4x^2-7x-15=0 \\ \\ using\text{ formula method} \\ \\ a=4 \\ b=-7 \\ c=-15 \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ x=\frac{7\pm\sqrt{49-4(4)(-15)}}{2(4)} \\ \\ x=\frac{7\pm\sqrt{289}}{8} \\ \\ x=\frac{7\pm17}{8} \\ \\ x=\frac{24}{8},\frac{-10}{8} \\ \\ x=3,-\frac{5}{4} \end{gathered}$

The x - intercept are

$\begin{gathered} x=3 \\ and \\ x=-\frac{5}{4} \end{gathered}$

f(x) vertex is a minimum

$\begin{gathered} f(x)=4x^{2}-7x-15 \\ \\ f(x)=4(x^2-\frac{7}{4}x)-15 \\ \\ f(x)=4(x-\frac{7}{8})^2-4(\frac{7}{8})^2-15 \\ \\ f(x)=4(x-\frac{7}{8})^2-4(\frac{49}{64})-15 \\ \\ f(x)=4(x-\frac{7}{8})^2-\frac{289}{16} \end{gathered}$

The coordinate of the vertex is

$(\frac{7}{8},-\frac{289}{16})$

The graph of the function is

Basically, we use the vertex point, intercept on the x - axis and y - axis