The solutions or roots so-called are really just the x-intercepts, and for a quadratic equation, the vertex is found right half-way between those x-intercepts.
we can get the intercepts by zeroing out "y" or f(x), so let's take a peek at one of those,

so, the x-intercepts or solutions are at 5 and -1, let's take a peek what's the halfway point.
-1----0----1----
2 ----3-----4------5
well then, now we know the vertex is at x = 2, but, what's the y-coordinate of it anyway?
y = (x - 5)(x + 1)
y = (2 - 5)(2 + 1)
y = (-3)(3)
y =
-9
f(x)=2x^2+0x-5
Find the vertex
0/4=0=x
Input x into the equation to find y or f(x)
0+0-5=f(x)
(0,-5) = Vertex
Now find a random x coordinate to input into the equation.
Let's do 4.
32+0-5=f(x)=27
It may not look like a parabola due to it being uneven.
Answer:
Step-by-step explanation:
(3-7)/(-2+1)= -4/-1= 4
y - 7 = 4(x + 1)
y - 7 = 4x + 4
y = 4x + 11
Answer:
18.
Step-by-step explanation:
If you divide by something less than 1, it'll wind up being more than what it was before.
The only one with two obtuse and two right angles.
Trapezoid