Given:
The function is
[tex]f(x)=4x^2-7x-15[/tex]Required:
Find the x-intercepts and describe the end behavior of the graph of f(x).
Explanation:
The given function is:
[tex]f(x)=4x^{2}-7x-15[/tex]To find the x-intercepts put f(x)=0
[tex]4x^2-7x-15=0[/tex]This is a quadratic equation solved by using the middle term splitting method.
[tex]\begin{gathered} 4x^2-12x+5x-15=0 \\ 4x(x-3)+5(x-3)=0 \\ (x-3)(4x+5)=0 \end{gathered}[/tex][tex]\begin{gathered} x-3=0 \\ x=3 \end{gathered}[/tex][tex]\begin{gathered} 4x+5=0 \\ x=-\frac{5}{4} \\ x=-1.25 \end{gathered}[/tex]Thus the x-intercept are: -1.25 and 3.
Compare the given equation with the standard equation
[tex]y=ax^2+bx+c[/tex][tex]a=4,b=-7,c=-15[/tex]The vertex is (h,k).
[tex]\begin{gathered} h=-\frac{b}{2a} \\ h=-\frac{-7}{2\times4} \\ h=\frac{7}{8} \\ h\approx0.9 \end{gathered}[/tex][tex]\begin{gathered} k=f(h) \\ k=f(0.9) \\ k=4(0.9)^2-7(0.9)-15 \\ k=3.24-6.3-15 \\ k\approx-18.1 \end{gathered}[/tex]Thus the vertex of the parabola is:
[tex](0.9,-18.1)[/tex]We will plot the points of the vertex and the x-intercepts and then draw the graph of the parabola as:
Here the leading coefficient a = 4>0
So the parabola opens upwards.
In the given function the degree is 2 which is even so as
[tex]\begin{gathered} x\rightarrow\infty,\text{ f\lparen X\rparen}\rightarrow\infty \\ x\operatorname{\rightarrow}\times-\infty,\text{f}\operatorname{\lparen}\text{X}\operatorname{\rparen}\operatorname{\rightarrow}\infty \end{gathered}[/tex]Final Answer:
As explained in the explanation part.
