How do I find the vertex of the parabola?

Question

y = = 5x^2 + 4x - 3

Answer 1

For a general form y= ax^2 +bx +c

The x coordinate of the vertex is x = -b/(2a)

The y coordinate can be found by substituting this x value into the first equation.


So in your case x = -4/(2*5) = -0.4
Then substitute -0.4 into y =5x^2 + 4x - 3

Answer 2

There are two approaches. In each case, you're going to find the x-coordinate of the vertex first, then plug that value into the equation to get the y-coordinate. What differs is how you go about finding the x-coordinate.

One approach is the one you probably learned in class: start by writing the parabola in the form y = ax² + bx + c (it may already look like that, but if not, simplify it until it does); at that point the x-coordinate will have the value -b/2a. In your case, that's -4/(2*5) = -2/5. (Plugging that value into the equation gives us the y-coordinate: y = 5(-2/5)² + 4(-2/5) - 3 = -19/5. )

The other approach won't work on this problem, but it does work on other problems, and it's good to keep in mind. It hinges on the fact that all parabolas are symmetrical, and in particular, when you can write your parabola as y = ax² + bx + c, then the "axis of symmetry" is a vertical line passing right through the vertex. That means that every point on the parabola on one side of the axis has a "mirror image" point on the other side of the axis, the same distance away.

This gives us a shortcut *if* the expression ax² + bx + c is easily factored: we take the average of the x-intercepts. Why? Because the x-intercepts are the two places where the parabola crosses the x-axis, and because the parabola is symmetrical, they must be mirror images of each other across the axis of symmetry. So the x-coordinate of the vertex will be midway between them (although we'll still need to plug that value into the equation to get the y-coordinate).

Now, the example you gave isn't factorable at all, so we'll use another example: y = x² + 3x -18, the right side of which can be factored as (x + 6)(x - 3). Obviously, this parabola will have a different vertex than the one you provided.

As you probably already know, when we set y = 0, we can solve what's left to get the x-intercepts -- the places where the curve crosses the x-axis.To solve 0 = (x + 6)(x - 3), we just use the Zero Product Law: either x + 6 = 0, which means x = -6, or x - 3 = 0, which means x = 3. In other words, the x-intercepts are (-6, 0) and (3, 0). Since the parabola is symmetrical, we can average those two vales (-6 + 3)/2 = -3/2 is the x-coordinate, and when we plug that into the equation, we get y = (-3/2)² + 3(-3/2) -18 = -81/4.

It's a shortcut method that doesn't work on every problem, but even if you never use it, understanding how it works will increase your understanding of parabolas, and graphing in general.

Hope that helps!

Answer 3

The function notation for the parabola is
y = ax² + bx + c
The basic notation is
(x - h)² = 4p(y - k)
Where h, k, and p are all numbers. If this is in this form, then the vertex is (h,k)

To solve for h and k,
y = ax² + bx + c
y = ax² + bx + c + b²/4a - b²/4a
y = ax² + bx + b²/4a + c - b²/4a
y = a(x² + b/a x + b²/4a²) + c - b²/4a
y = a(x + b/2a)² + c - b²/4a
y/a = (x + b/2a)² + c/a - b²/4a²
(x + b/2a)² = y/a - c/a + b²/4a²
[x - (-b/2a)]² = 1/a [y - (c - b²/4a)]

but,
[x - h]² = 4p[y - k]

so,
Therefore,
h = -b/2a
k = c - b²/4a
the vertex is
V( -b/2a , c - b²/4a )
You can always use this vertex formula so you will save effort and time
In your parabola
y = 5x² + 4x - 3
a = 5
b = 4
c = -3

h = -b/2a = -4/2(5) = -2/5
k = c - b²/4a = -3 - 4²/4(5) = -3 - 4/5 = -19/5
The vertex is
V(h,k)
= V( -b/2a , c - b²/4a )
= V(-2/5,-19/5)

^_^

Answer 4

y = 5x^2 + 4x - 3
y = (5x^2 + 4x) - 3
y = 5(x^2 + (4/5)x) - 3
y = 5(x^2 + (4/5)x + (4/25) - (4/25)) - 3
y = 5((x + (2/5))^2 - (4/25)) - 3
y = 5(x + (2/5))^2 - (20/25) - 3
y = 5(x + (2/5))^2 - (4/5) - 3
y = 5(x + (2/5))^2 - (4/5) - (15/5)
y = 5(x + (2/5))^2 - (19/5)

Vertex : ((-2/5),(-19/5))

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Unless you meant

y = -5x^2 + 4x - 3

then

y = -5(x - (2/5))^2 + (4/5) - (15/5)
y = -5(x - (2/5))^2 - (11/5)

Vertex : ((2/5),(-11/5))

Answer 5

compare your equation with

X^2 = 4AX

Here Vertex V (0,0)

first make quardatic equation into square one and get the answer

Answer 6

by using the formula y=a(x-h)squared by 2+k where (h,k) are the vertex

Answer 7

Try using the formula -b/2a for x then substitute to get y...
not sure though

Answer 8

Mine usual fall down the back of the settee. . . . . .