Anyone here have a math background?

[NotMeantForTheCity]

I was trying to explain to a student, whom I tutor, how to find the vertex of a parabola in order to explain symmetry. According to the Algebra text that I used in 8th grade the standard form for a parabolic equation is y = ax^2 + bx + c and the abscissa of the vertex = -{b/(2a)}

One of the exercise problems from this book is to find the vertex of the graph for y = x^2 - 4. But according to the book -4 is the b term rather than the c term. Can anyone explain why?

 

[Pearl]

I was trying to explain to a student, whom I tutor, how to find the vertex of a parabola in order to explain symmetry. According to the Algebra text that I used in 8th grade the standard form for a parabolic equation is y = ax^2 + bx + c and the abscissa of the vertex = -{b/(2a)}

One of the exercise problems from this book is to find the vertex of the graph for y = x^2 - 4. But according to the book -4 is the b term rather than the c term. Can anyone explain why?

If Neb sees this he can!!

 

[Neb]

I was trying to explain to a student, whom I tutor, how to find the vertex of a parabola in order to explain symmetry. According to the Algebra text that I used in 8th grade the standard form for a parabolic equation is y = ax^2 + bx + c and the abscissa of the vertex = -{b/(2a)}

One of the exercise problems from this book is to find the vertex of the graph for y = x^2 - 4. But according to the book -4 is the b term rather than the c term. Can anyone explain why?

for y = x^2 - 4.

There is no "x" so the b is zero. The term that doesn't include "x" is the constant "c".


Ben

 

[ClemKadiddlehopper]

I see that Neb has already answered the question, but here is my DH's explanation anyways.

The first equation is a quadratic equation. The special case where b=0 is commonly known as the parabolic equation. I suspect, when they start talking about parabolic equations there are only two constants, so the text book probably set the first constant to A and the second one to B which would be equal to C of the original quadratic equation.

 

[NotMeantForTheCity]

I misread the answer given in the book. The vertex is (0, -4). But, how can you get 0 as the value of the x coordinate by using the formula -{b/(2a)}? There's no b term in the problem's equation of y = x^2 - 4, and if the x value is 0, can any numbers be used in the formula to make the formula true?

 

[Fireman13]

If you were to write the given equation y = x^2 - 4 in the standard form it would be y = (1)x^2 + (0)x + (-4). The x term in the equation has to have a coefficient of zero in order to equal zero, so b = 0. Note, the b = 0 in the given equation, not the x.

Since b = 0, to complete the problem, you use x = (-b)/(2a) to solve for the x-coordinate which results in
x= (-)(0)/(2)(1)
x = 0/2
x = 0.

Substituting that back into the quadratic to find the y-coefficient you get,
y = (1)(0)^2 + (0)(0) + (-4)
y = 0 + 0 + (-4)
y = -4

Therefore the vertex coordinates are (0, -4)

 

[HippoTwilight]

I just want to take this opportunity to tell you all about Khan Academy. They have a website and a YouTube channel. Everything I've used was free, I think they operate on donations and ad revenue.

Pretty sure they cover lower level math too, but that guy got me thru 3 levels of calculus, differential equations, chemistry 2, and Intro to Controls Theory

I already have their entire YT channel downloaded to tutor my kid if they need it. Because I may be good at many STEM things, but I'm terrible at teaching it.

 

[NotMeantForTheCity]

If you were to write the given equation y = x^2 - 4 in the standard form it would be y = (1)x^2 + (0)x + (-4). The x term in the equation has to have a coefficient of zero in order to equal zero, so b = 0. Note, the b = 0 in the given equation, not the x.

Since b = 0, to complete the problem, you use x = (-b)/(2a) to solve for the x-coordinate which results in
x= (-)(0)/(2)(1)
x = 0/2
x = 0.

Substituting that back into the quadratic to find the y-coefficient you get,
y = (1)(0)^2 + (0)(0) + (-4)
y = 0 + 0 + (-4)
y = -4

Therefore the vertex coordinates are (0, -4)

I finally figured it out. I've been half-asleep for the past 4 days due to my blood pressure, and I had both an Algebra II and a pre-algebra student to deal with at the same time. I was thrown by having -4 in both the equation and the vertex.