Help in finding the x,y intercepts and T.P!!!!!!!!!!!

how do i find the x and y intercepts as well as the turning point!!!!

A) y=1.5 + 0.8x
B) y=2.8 + 0.7x
C) y=0.1x(x-4.5)
D) y=0.03t^2 + 0.6x

thank you soooooo much ^^

The x intercept is where the graph crosses the x axis when y = 0
So,

A) 1.5+0.8x = 0
0.8x = -1.5
x = -1.5/0.8 = -0.1875
intercept = (-0.1875, 0)

B) 2.8 +0.7x = 0
0.7x = -2.8
x = -2.8/0.7 = -4
intercept = (-4, 0)

C) 0.1x(x-4.5) = 0
x = 0 or x = 4.5
intercepts = (0,0) and (4.5, 0)

D) 0.03t^2+0.6t = 0 [Assuming it should read 0.6t]
t = 0 or t = 20
intercepts = (0,0) and (20,0)

The y intercept is where the graph crosses the y axis where x (or t in the case above) = 0
So,

A) y = 1.5+0.8(0) = 1.5
intercept = (0, 1.5)

B) y = 2.8 +0.7(0) = 2.8
intercept = (0, 2.8)

C) y = 0.1(0)(0-4.5) = 0
intercept = (0, 0)

D) y = 0.03(0)^2+0.6(0) =0
intercept = (0, 0)


You will only need to find a turning point if equation is quadratic i.e x^2 or above. i.e x^3 etc. In the case of quadratics, factorise the equation to find the roots, add these and divide by 2 to find the axis of symmetry. Plug this x value into the equation to find the y value and, hence the coordinates of the turning point. If the sign is + before the x^2 term the turning point is a minimum. If the sign is -, then it's a maximum.