Question

An expression is shown below:

f(x) = 2x2 − x − 10

Part A: What are the x-intercepts of the graph of f(x)? Show your work. (2 points)

Part B: Is the vertex of the graph of f(x) going to be a maximum or minimum? What are the coordinates of the vertex? Justify your answers and show your work. (3 points)

Part C: What are the steps you would use to graph f(x)? Justify that you can use the answers obtained in Part A and Part B to draw the graph. (5 points)

Answer 1

@Hero

Answer 2

You could graph this first

Answer 3

Part A :
The X-intercepts of the Graph of f(x) are the x-co-ordinates of the point/s of intersection of the Graph and the X-Axis.
Clearly, such a point/s is/are on the X-Axis, we must have, y=f(x)=0.
So, we solve \[f(x)=0, i.e., 2x ^{2}-x-10=0.\]
Now,\[2x ^{2}-x-10=0 \rightarrow (2x ^{2}-5x)+(4x-10)=0. \]
\[\rightarrow x(2x-5)+2(2x-5)=0.\]
\[\rightarrow (2x-5)(x+2)=0.\]
\[\rightarrow x=5/2, or, x=-2. \]
Hence, the desired intercepts are \[5/2, and, -2.\]

Answer 4

Part B :
To find the vertex of the graph of \[f(x)=y=2x ^{2}-x-10\], we proceed as under :\[y=2x ^{2}-x-10 \rightarrow y+10=2x^2-x.\]
Dividing by 2, we get, \[y/2+5=x^2-x/2\]
Now, to make \[x^2-x/2\] perfect square, we add, 1/16 on both sides.
\[:. y/2+5+1/16=x^2-x/2+1/16, i.e., y/2+81/16=(x-1/4)^2.\]
Multiplying by 2, we have,\[y+81/8=2(x-1/4)^2, or, y-(-81/8)=2(x-1/4)^2.\]
From this, we conclude that the vertex of the graph of \[f(x)\] is the point\[(1/4,-81/8).\]