F(x) = 2x2 − 3x − 5 part a: what are the x-intercepts of the graph of f(x)? show your work. 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. 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.

hello!

a. to find the x-intercepts of the graph of f(x), you would need to factor the equation.

2x² - 3x - 5 = 0

using the ac method, you get two factors: -5 and 2. since a > 1, you need to factor by grouping.

2x² - 5x + 2x - 5 = 0

(2x² - 5x) + (2x - 5) = 0 (find the gcf).

x(2x - 5) = 0 and 1(2x - 5) = 0

x + 1 = 0 and 2x - 5 = 0 you get x + 1 = 0 from the greatest common factors you factored out from grouping (2x² - 5x) and (2x - 5). if there was no gcf, then the factor is automatically 1.

x + 1 = 0 (subtract 1 from both sides)

x = -1

2x - 5 = 0 (add 5 to both sides)

2x = 5 (divide both sides by 2)

x = 5/2

part a: the x-intercepts of the graph are -1 and 5/2.

b. to find the vertex of the graph, use the formula x = -b/2a. then, you would substitute that value into f(x).

quadratics are formatted as ax² + bx + c = 0.

so then, you substitute the values from the given equation into the formula to find the vertex.

x = ) / 2(2)

x = 3/4

y = 2(3/4)² - 3(3/4) - 5

y = 9/8 - 9/4 - 5

y = -49/8

part b: the vertex of the graph is (3/4, -49/8), or (0.75, -6.125). since the function is not reflected over the x-axis, the vertex is a minimum value.

c. to graph f(x), the easiest way is to make a table of values, use the vertex, and the x-intercepts.

part c: you can use the answers obtained in part a and part b because if you were to substitute the x-values from part a and b into f(x), and simplify, they would have the same y-value.

i posted the graph of this equation, and table of values. i also included the points from the table of values onto the graph.

hopefully this !

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0 is its x and y coordinate