Quadratic equation: The vertex of y = ax^2 + bx – 5 is (3, 13). Find a and b.

[Nikolas111]

Quadratic equation: The vertex of y = ax^2 + bx – 5 is (3, 13). Find a and b.

Hello

The question is : The vertex of y = ax^2 + bx – 5 is (3, 13). Find a and b.



I found the first equation by subtituting (3,13) I got 13 =9a+3b-5 ->18=9a+3b I expressed from it a variable b and I got b=6-3a but I can't continue because I don't know how to create the second equation . Please could you help me . Thank you in advance

 

[stapel]

To start, please let me say "Thank you!" for showing your work and reasoning so nicely!

The vertex of y = ax^2 + bx – 5 is (3, 13). Find a and b.

I found the first equation by substituting (3,13) into the equation for (x, y). I got 13 = 9a + 3b - 5 => 18 = 9a + 3b. I expressed from it variable b in terms of a, and I got b = 6 - 3a.

But I can't continue because I don't know how to create the second equation.

By nature of parabolas, whatever value y is, some number of x-units to the left of the vertex, it's the same value at the same number of x-units to the right of the vertex. So let's go, say, one unit to the left and to the right (so the inputs will be 2 and 4). This gives us:

. . . . .\(\displaystyle 13\, =\, 4a\, +\, 2b\, -\, 5\, =\, 16a\, +\, 4b\, -\, 5\)

The "13=" part is actually not needed here. So let's simplify:

. . . . .\(\displaystyle 4a\, +\, 2b\, -\, 5\, =\, 16a\, +\, 4b\, -\, 5\)

Eliminate the "-5" from either side by adding:

. . . . .\(\displaystyle 4a\, +\, 2b\, =\, 16a\, +\, 4b\)

. . . . .\(\displaystyle 0\, =\, 12a\, +\, 2b\)

You already have:

. . . . .\(\displaystyle b\, =\, 6\, -\, 3a\)

So let's plug this in:

. . . . .\(\displaystyle 0\, =\, 12a\, +\,2(6\, -\, 3a)\, =\, 12a\, +\, 12\, -\, 6a\, =\, 6a\, +\, 12\)

. . . . .\(\displaystyle -12\, =\, 6a\)

. . . . .\(\displaystyle -2\, =\, a\)

Can you see where to go from here?

 

[Nikolas111]

Quadratic equation: The vertex of y = ax^2 + bx – 5 is (3, 13). Find a and b.

I understand almost everything just I don't know why exactly 2 and 4 are inputs could you explain me it please Thank you for helping me

 

[ecoo]

By nature of parabolas, whatever value y is, some number of x-units to the left of the vertex, it's the same value at the same number of x-units to the right of the vertex. So let's go, say, one unit to the left and to the right (so the inputs will be 2 and 4). This gives us:

. . . . .\(\displaystyle 13\, =\, 4a\, +\, 2b\, -\, 5\, =\, 16a\, +\, 4b\, -\, 5\)

The "13=" part is actually not needed here. So let's simplify:
.

You don't end up using the 13, but I don't see how 13 has any relation with the equations.

I understand almost everything just I don't know why exactly 2 and 4 are inputs could you explain me it please Thank you for helping me

The vertex is centered on x=3, and stapel already mentioned that points to the left of the vertex have an equal point on the right of the vertex, in terms of the y-value. In this case, 2 and 4 are chosen because the two points are one interval away from the vertex. Any two points that are of equal distance from the vertex can also be chosen, such as 1 and 5.

 

[Deleted member 4993]

Hello

The question is : The vertex of y = ax^2 + bx – 5 is (3, 13). Find a and b.



I found the first equation by subtituting (3,13) I got 13 =9a+3b-5 ->18=9a+3b I expressed from it a variable b and I got b=6-3a but I can't continue because I don't know how to create the second equation . Please could you help me . Thank you in advance

The vertex of a parabola is at x = -b/(2a) → 3 = - b/(2a) → 6a + b = 0 (2nd. equation)