# A parabola has a y-intercept of 3 and passes through the points (-2,15) and (4,3). What are the coordinates of the vertex?

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### 2 Answers

A parabola is a graph of a quadratic function. The equation of a quadratic function in standard form is

`y = ax^2 + bx + c`

The vertex of the parabola can be determined from this equation, so first we have to find a, b, and c, in order to write the equation of the given parabola.

The y-intercept of a parabola, or any graph, is the point where the graph intersects y-axis. At this point x = 0. So, for a quadratic function, y-intercept is always y = c.

Since the given parabola has the y-intercept of 3, c = 3. The equation is then

`y = ax^2 + bx + 3`

The two points on the parabola are given, so we can plug in their coordinates into the equation above in order to find a and b.

Point (-2, 15): x = -2, y = 15

`15 = a(-2)^2 + b(-2) + 3`

15 = 4a - 2b + 3

12 = 4a - 2b

Dividing both sides by 2 yields

6 = 2a - b

Point (4, 3): x = 4, y = 3

`3 = a*4^2 + b*4 + 3`

3 = 16a + 4b + 3

Subtracting 3 from both sides and dividing by 4 yields

4a + b = 0

These two equations can be solved together as a system:

2a - b = 6

4a + b = 0

This system can be solved using elimination. Adding the equations together results in

6a = 6

a = 1, which means b = -4a = -4

So the equation of the given parabola is

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

The x-coordinate of the vertex is determined by the formula

`x= -b/(2a)`

Plugging in b = -4 and a = 1, we get `x = -(-4)/(2*1) = 2` .

The y-coordinate of the vertex can be found by plugging the x-coordinate into the equation of the parabola:

`y = 2^2 -4*2 + 3 = 4 - 8 + 3 = -1`

**Thus, the vertex of the given parabola is located at the point (2, -1).**

**x-coordinate is 2 and y-coordinate is -1.**

We know that a parabola is the graph for the quadratic function. As such, it has the general equation f(x)= y = ax^2 + bx + c

The y-intercept occurs when the value of x is 0. Therefore, at the y-intercept,

y = c

but we know that the y-intercept is (0,3) and therefore

y = 3

the equation of the parabola now becomes

y = ax^2 +bx+3

plugging in for the two points given (i.e. (-2,15) and (4,3)) then we get two equations that can be solved simultaneously for coefficients a and b:

15 = 4a - 2b + 3 (1)

3 = 16a + 4b+ 3 (2)

adding equation (2) to two times equation (1) gives

30+3=(8+16)a+(-4+4)b+(6+3)

solving for a

33-9=24=24a

==> a=1

to solve for b, plug a+1 into equation (1)

15 = 4(1) -2b+3

2b= 8

==> b= -4

therefore the quadratic equation is

f(x) = y = x^2 -4x +3

The vertex can now be found.

dy/dx = 2x - 4

when dy/dx = 0 we have the vertex

therefore, 2x-4= 0

==> x=2

==> y= (2)^2 -4(2) +3

==> y= -1

therefore, the vertex of the given parabola occurs at (2,-1)