Find b and c so that y=−17x^2+bx+c has vertex (−1,5)

Question

Find b and c so that y=−17x^2+bx+c has vertex (−1,5)

Answer 1

Hello Alishia!

So we need to find the values of b and c which make that parabola have a vertex at (-1, 5).

If we try plugging in -1 for x and 5 for y in the above equation, we get:

5 = -17×(-1)² + b×(-1) + c

But we cannot solve this, since we have 1 equation and 2 unknowns.

If there are two variables you don't know the solution to, you need 2 equations.

Luckily we have another equation we can use.

Are you familiar with the quadratic formula?

The quadratic formula gives us the X value for each X intercept.

given a quadratic in the general form:

y = ax² + bx + c, we can find the x-intercepts (where it crosses the x axis (i.e. where y = 0) ) with the following formula:

x = (-b +- sqrt(b² - 4ac))/2a

Notice the + or - in the middle there.

There are usually 2 solutions to a quadratic.

Take for example, y = x² - 4

Y= 0 at x =2 and at x=-2, so both of these values are places where the graph crosses the x-axis.

If you think about it, both solutions are equally far from the vertex, since parabolas are symmetric.

In the above example, the vertex was at 0, and the two solutions were at -2, and 2.

Each 2 from the vertex.

Therefore, we can take both solutions to the quadratic equation, add them together, and divide by 2 to get the x-coordinate of the vertex.

This works because adding two numbers together and dividing by 2 takes the average of the two numbers, which ends up right in the middle of them both.

(For example: (3 + 5)/2 = 8/2 = 4, which is indeed right in the middle)

So, that would entail:

[ (-b + sqrt(b² − 4ac))/2a + (-b − sqrt(b² − 4ac))/2a ] / 2

(notice that we are dividing the entire thing by 2 in order to find the midpoint)

the sqrt parts will cancel out, leaving:

[(-b −b)/2a]/2 => -[2b/2a]/2 => (-b/a)/2 => -b/2a

So the x-coordinate of the vertex is at -b/2a.

In our above example with y = x^2 -4, we could rewrite the equation as:

y = 1x² + 0x - 4

Thus b = 0, and a = 1

Yielding a vertex with x at -0/2, which is 0, as expected.

So now we have a second equation.

We know the x-coordinate of the vertex must be -1.

So, we have:

-1 = -b/2a.

Well... b is unknown, we are trying to solve for it.

But we know a. In our example, a is -17.

(y = -17x^2 +bx + c, and a is the coefficent of x^2)

So our second equation is:

-1 = -b / 2*(-17).

Multiplying both sides by 2 * (-17) yields:

-b = -1 * 2 * (-17)

Dividing by -1 yields:

b = 2 * -17

b = -34

Then we can plug b back into the equation from our original attempt :

First equation: 5 = -17×(-1)² + b×(-1) + c

and can solve for c.

I hope this helps! If you don't understand something, don't hesitate to message me or to reply here!