Quadratic function problem.?
find the rule of correspondence for a quadratic function whose graph has the indicated features. Write the rule in standard form.
1. x- intercepts at (-3,0) and (1,0); the same shape as y=x^2
2.x- intercepts at (7/2,0) and (5,0); the same shape as y= -3x^2
what do you mean this is straight out of the textbook the question, I'm trying to figure out how you don't understand what rule of correspondence means.
- Ray SLv 79 years agoFavorite Answer
➊ For y = ax² + bx + c
If h and k are zeros of the quadratic function, then (x-h) and (x-k)
are factors of it ... zeros being the value(s) of x that make
the function zero, i.e. the value(s) of x that produce y=0.
Notice that this doesn't allow for all coefficients 'a' of x²:
➋ y = (x-h)(x-k) = x² -hx - kx + hk = x² + (-h-k)x + hk ← b = (-h-k)
c = hk
But, 'a' can only be 1.
So, we need to incorporate 'a' into ➋.
That modifies ➋ to the final form
➌ y = a(x-h)(x-k)
Now, using ➌, your problems become easy to do.
1. x-intercepts at (-3,0) and (1,0); the same shape as y = x².
x-intercepts are the zeros of the function ... i.e. 1 & -3 are zeros
and the same shape as y = x² indicate that a=1.
➌ y = a(x-h)(x-k) = 1(x-1)(x-(-3)) = (x-1)(x+3) = x²+2x-3
y = x²+2x-3 ← ANSWER
2. x-intercepts at (7/2,0) and (5,0); the same shape as y= -3x²
x-intercepts are ⁷∕₂ & 5 and a = -3
➌ y = a(x-h)(x-k) = -3(x-⁷∕₂)(x-5) = (x-⁷∕₂)(x-5) = -3x²+(⁵¹∕₂)x - ¹⁰⁵∕₂
y = -3x²+(⁵¹∕₂)x - ¹⁰⁵∕₂ ← ANSWER
Hope this is what you are looking for.
Have a good one!