Question

Complete the square and find the x and y intercepts.

y=-2x^2+3x+4

Answer 1

To complete the square, you first make the x^2 "alone", that is to say, it must have 1 as its coefficient :)

Answer 2

I got -2(x-3/4)+41/9

Answer 3

y-intercept = 4
x-intercept=2.35 and -0.851

Answer 4

Nicely done, more quickly than I did, :D

Answer 5

@sauravshakya could you show me your work?

Answer 6

for y-intercept, x=0 so,
y =-2*(0)^2+3*(0)+4=4
thus,4 is the y-intercept
for x intercept, y=0 so,
0=-2x^2+3x+4
or,2x^2-3x-4=0
Now use any method to solve the above eqn to get the value of x.
And this values of x will be x-intercept.

Answer 7

complete the square
\[ -2(x^2 + -\frac{3}{2}x -2) = y \]
multiply the coeff of x by 1/2 and then square it: -3/2 * 1/2 = -3/4
squared: 9/16
add and subtract 9/16 to the stuff inside the parens (9/16-9/16=0 so we are not changing its value):
\[ -2(x^2 + -\frac{3}{2}x +\frac{9}{16}-\frac{9}{16}-2) = y \]
group the first 3 terms inside the parens:
\[ -2( (x^2 + -\frac{3}{2}x +\frac{9}{16})-\frac{9}{16}-2) = y \]
the grouped terms are a perfect square:
\[ -2( (x - \frac{3}{4})^2-\frac{9}{16}-2) = y \]
combine the last 2 terms in side the parens:
\[ -2( (x - \frac{3}{4})^2-\frac{41}{16}) = y \]

Answer 8

to find the x intercepts, set y equal to zero
\[ -2( (x - \frac{3}{4})^2-\frac{41}{16}) = 0 \]
divide through by -2 to simplify
\[ (x - \frac{3}{4})^2-\frac{41}{16} = 0 \]
move the constant to the right side
\[ (x - \frac{3}{4})^2=\frac{41}{16} \]
take the square root of both sides
\[ x - \frac{3}{4}=\frac{\pm\sqrt{41}}{4} \]
add 3/4 to both sides
\[x = \frac{1}{4}(3\pm\sqrt{41}) \]