Question

Please show detailed work if possible-that will help me to better understand the questions

Answer 1

Answer:

We are given the function:

$f(x)=2x^2-x-10$

$Here,\\a=2, b=-1,c=-10$

1. X-intercepts are the points at which the graph of a function intersects or cuts the x-axis. Since the x-intercept always lies on the x-axis, its ordinate or y-coordinate will always be 0. Since the function is quadratic, it will have at most 2 x-intercepts.

In order to find the x intercept, we basically solve for x at y=0:

$f(x)=2x^2-x-10\\As\ y=0,\\0=2x^2-x-10\\2x^2-x-10=0\\ 2x^2-5x+4x-10=0\\x(2x-5)+2(2x-5)=0\\(x+2)(2x-5)=0\\Hence,\\Individually:\\x=-2,\ x=\frac{5}{2}$

Hence, the x-intercepts of the parabola of f(x) is (-2,0),(2.5,0)

2. The vertex of parabola is determined as maximum or minimum, solely on how it opens. This depends on the nature of the co-efficient of the x^2 term or 'a'. If a is positive the parabola opens upwards (minimum point) and downwards (maximum point) if negative. Hence, here as a=2, the parabola opens upwards and its vertex is minimum.

$Vertex=(\frac{-b}{2a},\frac{-D}{4a})\\Hence,\\D=b^2-4ac\\Substituting\ a=2,b=-1,c=-10:\\D=(-1)^2-4*2*-10=1+80=81\\Hence,\\Vertex\ of\ f(x)=(\frac{-(-1)}{2*2},\frac{-81}{4*2})=(\frac{1}{4},\frac{-81}{8})$

3. [Please refer to the attachment]

From the graph, we observe that the parabola cuts the x-axis at (-2,0),(2.5,0). Also, its clear that the axis of symmetry passes through $(\frac{1}{4},\frac{-81}{8})$, which is its minimum point.