Which of the following is true for the graph of the quadratic function f(x)=2x^2-10x+12
a. the vertex of the function is the point (5/2), (-1/2)
b. the y-intercept is the point (0, 10).
c. the x-intercept is the point (3, 2).
d. all of the above

To find the x-intercepts let f(x) = 0 and solve the equation. To find the y-intercept(s) let x = 0 and solve the equation. To find the vertex: the x coordinate of the vertex is -b/2a where a is the coefficient of the x^2 and b is the coefficient of x. Then, substitute the value you get for -b/2a for the x and solve for y. That will give you the x,y coordinates. 
4. a. Solve 2x^2 -10x + 12 =0. Either use the quadratic formula or factor. x = 2 and x = 3. But those are the intercepts. The points of the intercepts are (2,0) and (3,0), NOT the point (3,2). a cannot be true. 4b. In the equation, substitute 0 for x and solve. f(x) or y comes out to 12. b is correct. 4c. -b/2a = 10/4 which reduces to 5/2. Substitute 5/2 for x and evaluate to find y. y = -1/2. Your answers for 4 are b and c. 
5. a. Substitute 0 for x and solve. y = -6, so your y-intercept is (0.-6). a is correct b. Substitute 0 for y and solve. x = 2 and -1. b is correct. That means d is not. c. -b/2a = 3/6 = 1/2. Substitute 1/2 for x and solve for y. The vertex is (1/2, - 27/4), e is correct. Your answers for 5 are a,b, and e. 
7. Substitute 0 for x and solve for y. y = -16/5, so the y intercept is (0/-16/5) a is true. b. x-intercepts always have 0 for the y coordinate, so b is NOT true. c. Expand your binomial and put your equation in ax^2, + bx + c form. -b/2a is 6/5 divided by 3/5 which equals 3. Put 3 in for x and find y. Your answer should be -5, so c is true. d. Let f(x) =0. Solve you equation by using the quadratic formula or factoring. x comes out to be 8 and -2, so d is true. Your ansers for 7 are a,c, and d.

Idon’t know i’m in the same boat