Quadratics Please Help?

Question

Hello,

There are some questions that I am not sure how to do:

1) Given that f(x) = 15 - 7x - 2x^2

1a) Find the coordinates of all the points at which the graph of y = f(x) crosses the coordinate axes.

 1b) Sketch the graph of y = f(x)

2) Given that for all values of x 3x^2 + 12x + 5 = p(x + q)^2 + r

2a) Find the values of p, q and r.

2b) Hence or otherwise find the minimum values of 3x^2 + 12x + 5

2c) Solve the equation 3x^2 + 12x + 5 = 0.

Thank you very much in advance.

Answer 1

1) The quadratic equation f(x) = 15 - 7x - 2x²

y = - 2x² - 7x + 15

1a) x intercepts and y intercepts are crosses the coordinate axis.

To find x intecept substitute y = 0 in quadratic function.

0 = - 2x² - 7x + 15

- 2x² - 7x + 15 = 0

- 2x² - 10x + 3x + 15 = 0

- 2x( x + 5) + 3(x + 5) = 0

(x + 5)(- 2x + 3) = 0

x = -5 and -2x + 3 = 0

x = -5 and x = 3/2

x intercepts are (-5,0) (3/2 , 0)

To find y intecept substitute x = 0 in quadratic function.

y = - 2(0)² - 7(0) + 15

y = 15

y intercept is (0,15)

The coordinates are (-5,0) (3/2 , 0) and (0,15)

1b) Graph of y = - 2x² - 7x + 15

Compare it to standard form of parabola y = ax² + bx + c

Axis of symmetry x = -b/2a = -7/4

To find vertex substitue x = -7/4 in y = - 2x² - 7x + 15.

y = - 2(-7/4)² - 7(-7/4) + 15

y = -2(49/16) + 49/4 + 15

y = - 98/16 + 49/4 + 15

y = (- 98 + 196 + 240)/16

y = 21.125

Vertex is (-7/4 , 21.125)

Choose integer values for x and evaluate the function for each value.

x	y = - 2x2 - 7x + 15	(x, y)
-6	y = - 2(-6)2 - 7(-6) + 15 = -15	(-6, -15)
-4	y = - 2(-4)2 - 7(-4) + 15 = 11	(-4, 11)
2	y = - 2(2)2 - 7(2) + 15 = -7	(2, -7)
3	y = - 2(3)2 - 7(3) + 15 = -24	(3, -24)
1	y = - 2(1)2 - 7(1) + 15 = 6	(1, 6)

Graph :

Draw the coordinate plane.

Plot the points found in the table.

Plot the x , y intercepts and axis of symmetry, vertex.

Connect the points with a smooth curve.

Answer 2

2) The equation 3x² + 12x + 5

y = 3x² + 12x + 5

2a) To find  the values of p, q, r

y = 3x² + 12x + 5

Standard form of parabola y  = ax ² + bx + c

Vertex form of parabola is y  = a (x - h ) ² + k

Here x ² coefficient is 3 , for perfect square make x ² coefficient 1 by dividing each side by 3.

y/3 = x² + 4x + 5/3

To complete the squre add (half cooefficient of x ) ² to each side of the expression.

x  coefficient = 4 then (half the x coefficient)² is (2) ² = 4.

y/3 + 4 = x² + 4x + 4 + 5/3

y/3 = (x + 2)² + 5/3 - 4

y/3 = (x + 2)² -7/3

 y = 3 (x + 2)² - 7

Now the equation is 3(x + 2)² - 7

Compare it to p(x + q)² + r

p = 3 , q = 2 and r = - 7

2b) Minimum value of quadratic function(parabols) is vertex.

The equation 3x² + 12x + 5

Vertex form of the equation is y = 3 (x (- 2))² - 7

y  = a (x - h )² + k

Where (h ,k) is vertex.

Vertex is minimum point (h ,k ) = (- 2, -7)

2c) 3x² + 12x + 5 = 0

Compare it to quadratic equation

Roots are

Solutions are