Find the following to accurately graph the quadratics below: a) Table

elvishwitchxyp

elvishwitchxyp

Answered question

2021-12-31

Find the following to accurately graph the quadratics below:
a) Table of 5 values (leave any non-whole numbers as fractions if not a terminating decimal)
b) The vertex (tell whether max/min.)
c) The y- intercept
y=−x2−8x−6

Answer & Explanation

Virginia Palmer

Virginia Palmer

Beginner2022-01-01Added 27 answers

Step 1
Since the given y=−x2−8x−6
for x-intercepts, y=6
−x2−8x−6=0=1x2+8x+6=0
(ax2+bx+c=0)
(x=−b±b2−4ac2a)
x=−8±64−242
x=−8±402
x=−8+402
−8−402=1x=−8+6.3242
−8−6.3242
x=−0.838,−7.162
So, curse cuts x-axis at (−0.838, 0) at (−7.162, 0)
for y-intercepts x=0
y=−6
So, curse cuts y-axis at (0, −6)
y=−(x2+8x+6)=−(x2+2×x×4+(4)2−(4)2+6)
y=−((x+4)2−16+6)=−((x+4)2−10)×−(x+4)2+10
y=a(h−h)2k
So, h=−4, k=10
So vertex is (h, k)=(−4, 10)
Now, make table
x246810−2−4−6−8−10y−26−54−90−18461066−26
Ronnie Schechter

Ronnie Schechter

Beginner2022-01-02Added 27 answers

Step 1
Given y=−x2−8x−6
Swap sides so that all variable terms are on the left hand side.
−x2−8x−6=y
Subtract y from both sides.
−x2−8x−6−y=0
All equations of the form ax2+bx+c=0 can be solved using the quadratic formula: −b±b2−4ac2a. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
−x2−8x−y−6=0
This equation is in standard form: ax2+bx+c=0. Substitute -1 for a, -8 for b, and −6−y for c in the quadratic formula, −b±b2−4ac2a
x=−(8)±(−8)2−4(−1)(−y−6)2(−1)
Square -8.
x=−(−8)±64−4(−1)(−y−6)2(−1)
Multiply -4 times -1
x=−(−8)±64+4(−y−6)2(−1)
Multiply 4 times −6−y
x=−(8)±64−4y−242(−1)
Add 64 to −24−4y
x=−(−8)±40−4y2(−1)
Take the square of 40−4y
x=−(−8)±210−y2(−1)
The opposite of -8 is 8
x=8±210−y2(−1)
Multiply 2 times -1
x=8±210−y−2
Now solve the equation x=8±210−y−2 when ± is plus. Add 8 to 210−y
x=210−y+8−2
Divide 8+210−y by -2
x=−(10−y+4)
When ± is minus. Subtract 210−
karton

karton

Expert2022-01-09Added 613 answers

Step 1
x-axis interception points of −x2−8x−6:(−4−10, 0), (−4+10, 0)
y-axis interception point of x2−8x−6:(0, −6)
Step 2
The vertex of an up-down facing parabola of the form y=ax2+bx+c is xv=−b2a
The parabola params are:
a=−1, b=−8, c=−6
xv=−b2a
xv=−(−8)2(−1)
Simplify
xv=−4
Plug in xv=−4 to find the yv value
yv=10
Therefore the parabola vertex is
(-4,10)
If a<0, then the vertex is a maximum value
If a>0 then the vertex is a minimum value
a=-1
Maximum (-4, 10)
 

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