QUADRATIC EQUATONS PRACTICE QUESTIONS FOR SAT

Problem 1 :

The function f is defined by f(x) = x² + bx + c where b and c are constants. If the graph of f has x-intercepts at -5 and 3. Which of the following correctly gives the values of b and c ?

a) b = -5, c = 3      b)  b = -3, c = 5      c)  b = -2, c = - 15

d)  b = 2, c = -15

Solution :

f(x) = x² + bx + c

Since the x-intercepts are -5 and 3, we can write it as (-5, 0) and (3, 0).

(1) - (2)

-5b + c - (3b + c) = -25 + 9

-8b = -16

b = 2

Applying the value of b in (2), we get

3(2) + c = -9

c = -9 - 6

c = -15

So, option d is correct.

Problem 2 :

y = x² - 2x - 3

The parabola in the xy-plane is given by the equation above. Which of the following equivalent forms of the equation displays the coordinate of the vertex of the parabola as constants or coefficients ?

a) y = (x - 1)² - 4            b) y = (x - 1)² - 2

c) y = (x - 3)(x + 1)        d)  y + 3 = x(x + 2)

Solution :

y = x² - 2x - 3

To express the quadratic function in vertex form, we have to write it in the form of y = a (x - h)² + k

y = x² - 2x(1) + 1² - 1² - 3

y = (x - 1)² - 1 - 3

y = (x - 1)² - 4

So, option a is correct.

Problem 3 :

y = x² + 10x + 16

The equation above represents a parabola in the xy-plane. Which of the following equivalent forms of the equation displays the minimum value of y as a constant or coefficient ?

a) y = (x + 8) (x + 2)       b)  y - 16 = x(x + 10)

c)  y = (x + 5)² - 9        d)  y = (x - 5)² + 9

Solution :

y = x² + 10x + 16

y = x² + 2 x (5) + 5² - 5² + 16

y = (x + 5)² - 25 + 16

y = (x + 5)² - 9

So, option c is correct.

Problem 4 :

y = x² - 10x + k

In the equation above, k is constant. If the equation represents a parabola in the xy-plane that is tangent to the x-axis, what is the value of k ?

Solution :

Since x-axis is the tangent of the parabola, the curve will touch x-axis at one point. It has one solution.

b² - 4ac = 0

a = 1, b = -10 and c = k

(-10)² - 4(1) (k) = 0

100 - 4k = 0

4k = 100

k = 25

Problem 5 :

A parabola is shown in the xy-plane above. Which of the following equations correctly represents the parabola by displaying the x-intercepts of the parabola as constants or coefficients ?

a)  y = (x + 1.5)² - 20.25        b)  y = (x - 1.5)² - 20.25

c)  y = (x + 6)(x - 3)           d)  y = (x - 6)(x + 3)

Solution :

y = a(x - h)² + k

Here (h, k)==>(-1.5, -20.25)

y = a(x - (-1.5))² + (-20.25)

y = a(x + 1.5)² - 20.25

Apply (-6, 0), we get

0 = a(-6 + 1.5)² - 20.25

0 = a(-4.5)² - 20.25

20.25a = 20.25

a = 1

y = 1(x + 1.5)² - 20.25

So, option a is correct.

Problem 6 :

The graph of the equation y = -x² + 6x + 16 is a parabola with (3, 25) as shown in the xy-plane above. If one of the x-intercepts is at -2, which of the following equivalent forms of the equation shows the x-intercepts of the parabola as constant or coefficients ?

a) y = -2(x + 2)(x - 8)      b) y = -(x + 2)(x - 8)

c)  y = (x + 2)(x - 8)        d)  y = -(x - 3)² + 25

Solution :

Converting the given into vertex form, we get

y = -[x² - 6x - 16]

y = -[x² - 2x(3) + 3² - 3² - 16]

y = -[(x - 3)² - 9 - 16]

y = -[(x - 3)² - 25]

y = -(x - 3)² + 25

Problem 7 :

In the xy-plane, the graph of a parabola has x-intercepts at -3 and 5. If the y-coordinate of the vertex of the parabola is 8, which of the following could be the equation of parabola ?

Solution :

x-intercepts are -3 and 5.

x = -3 and x = 5

y = a(x + 3)(x - 5)

Midpoint of x-intercepts = x-coordinate of vertex

h = (-3 + 5)/2

h = 2/2

h = 1 and k = 8 (given)

Vertex is also one of the points on the parabola.

8 = a(1 + 3)(1 - 5)

8 = a(4)(-4)

-16a = 8

a = -1/2

So, the required equation is,

y = (-1/2)(x + 3)(x - 5)

Problem 8 :

h = -6t² + 36t + 12

The height of a model rocket is modeled by the equation above, where h is the height of the rocket, in meters and t is the number of seconds after launch. In which of the following equations the number of seconds it takes the rocket to reach the maximum height appear as constant or coefficients ?

a) h = -6(t +3)² + 42       b)  h = -6(t - 3)² + 66

c)  h = -6(t² - 6t - 2)        d)  h = -6(t - 2)(t - 4) + 60

Solution :

h = -6t² + 36t + 12

h = -6[t² - 6t - 2]

h = -6[t² - 2(t)(3) + 3² - 3² - 2]

h = -6[(t - 3)² - 9 - 2]

h = -6[(t - 3)² - 11]

h = -6(t - 3)² + 66

So, option b is correct.