Show all an expression is shown below: f(x) = 2x2 − 5x + 3 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)

We have that2x²  − 5x + 3we know thatthe  x-intercepts of the graph of f(x) is when y=0then  2x²  − 5x + 3=> resolving the second order equationx=1x=3/22x²  − 5x + 3=(x-1)*(x-3/2)  the answer part a) isx=1x=3/2using a graph toolsee the attached figurepart b: is the vertex of the graph of f(x) going to be a maximum or minimum? what are the coordinates of the vertex? the graph has a minimum, because a, the coefficient of x², is a positive number 2, so the graph opens upward, the graph has a lowest point, which is the minimum.the vertex is when x=-b/2a=)/(2*2)=5/4when x=5/4, f(5/4)=2(5/4)²-5(5/4)+3=-1/8so the coordinates of the vertex is (5/4, -1/8)part c: what are the steps you would use to graph f(x)?   to graph a parabola, you need to have five points. you already have three: the two x intercepts and the vertex. to get another two points, set x=0, f(x) is then 3, so you have the point (0,3).  notice x=5/4 is the line of symmetry, so from 0 to the symmetry line x=5/4 the distance is 5/4, add 5/4 to the right side 5/4 +5/4=5/2. when x=5/2, f(x) also=3. locate the point (5/2, 3)connect all the five points, you have your graph