SOLUTION: Please help me with these problems: Finding the maximum y-value on the graph of y=f(x). My problem is f(x)= -x^2+8x+7 Using quadratic formula to find any x-intercepts on gr

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help me with these problems: Finding the maximum y-value on the graph of y=f(x). My problem is f(x)= -x^2+8x+7 Using quadratic formula to find any x-intercepts on gr      Log On


   

Question 724015: Please help me with these problems:
Finding the maximum y-value on the graph of y=f(x).
My problem is f(x)= -x^2+8x+7
Using quadratic formula to find any x-intercepts on graph of equation. My problem is y=x^2+6x-1.
and
To solve by completing the square what value should be added to each side of equation with problem x^2+16x=-4.
Thanking you in advance for your help.
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Finding the maximum y-value on the graph of y=f(x).
My problem is f(x)= -x^2+8x+7
Vertex occurs at x = -b/(2a) = -8/(2*-1) = 4
Max at f(4) = -16+32+7 = 16+7 = 23
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Using quadratic formula to find any x-intercepts on graph of equation. My problem is y=x^2+6x-1.
Let y = 0
x^2+6x-1 = 0
x = [-6+-sqrt(36-4*-1)]/2
x = [-6+-sqrt(40)]/2
x = [-3+-sqrt(10)]
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and
To solve by completing the square what value should be added to each side of equation with problem x^2+16x=-4.
x^2+16x+64 = -4+64
-----
(x+8)^2 = 60
x+8 = +-2sqrt(15)
x = -8+-2sqrt(15)
=====================
Cheers,
Stan H.
=====================

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
1. f%28x%29=+-x%5E2%2B8x%2B7...here you have parabola with a=-1, so parabola is facing downward and the maximum y-value on the graph will be y coordinate of the vertex
the equation is y+=+%28x+-+h%29%5E2+%2B+k where h and k are x and y coordinate of the vertex
so, we need k
f%28x%29=+-x%5E2%2B8x%2B7...factor completely
f%28x%29=+-x%5E2%2B8x%2B7...replace 7 with -16%2B23
f%28x%29=-x%5E2%2B8x-16%2B23...group first three terms together
f%28x%29=-%28x%5E2-8x%2B16%29%2B23.......note that -%28x%5E2-8x%2B16%29=-%28x-4%29%5E2
f%28x%29=-%28x-4%29%5E2%2B23...vertex form...h=4 and k=23
so, vertex is at 4%2C23 and the maximum y-value on the graph is y=23
see it on a graph:


2.
Using quadratic formula to find any x-intercepts on graph of equation. +y=x%5E2%2B6x-1

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ ...note that a=1, b=6, and c=-1

x+=+%28-6+%2B-+sqrt%28+6%5E2-4%2A1%2A%28-1%29+%29%29%2F%282%2A1%29+
x+=+%28-6+%2B-+sqrt%28+36%2B4+%29%29%2F2+
x+=+%28-6+%2B-+sqrt%28+40+%29%29%2F2+
x+=+%28-6+%2B-+6.3%29%2F2+
solutions:
x+=+%28-6+%2B+6.3%29%2F2+
x+=+0.3%2F2+
x+=+0.15+
and
x+=+%28-6+-6.3%29%2F2+
x+=+-12.3%2F2+
x+=+-6.15+
so, x-intercepts are at (0.15,0) and (-6.15,0)
check it on a graph:




3. complete the square
x%5E2%2B16x=-4
x%5E2%2B16x%2B4=0...replace 4 with 64-60
x%5E2%2B16x%2B64-60=0...group first three terms together
%28x%5E2%2B16x%2B64%29-60=0....note that %28x%5E2%2B16x%2B64%29 is square of a sum
%28x%2B8%29%5E2-60=0