Parabola Questions and Problems with Detailed Solutions

The x-intercepts are the intersection of the parabola with the x-axis, which are points on the x-axis and therefore their y-coordinates are equal to 0. Therefore, we must solve the equation: \[ 0 = - x^2 + 2x + 3 \] Factor the right side of the equation: \[ -(x - 3)(x + 1) = 0\] Solve for x to find: \[ x = 3 \quad \text{and} \quad x = -1\] , The y-intercepts are the intersection of the parabola with the y-axis, which is a point on the y-axis and, therefore, its x-coordinates equal 0 the y-intercept is :\( y = - (0)^2 + 2 (0) + 3 = 3 \), The vertex is found by writing the equation of the parabola in vertex form \(y = a(x - h)^2 + k \) by completing the square and identifying the coordinates of the vertex \( h \) and \( k \) . Complete the square: \[ y = - x^2 + 2 x + 3 = -( x^2 - 2 x - 3) = -( (x - 1)^2 - 1 - 3) = -(x - 1 )^2 + 4 \] Vertex at point \[ (1 , 4) \] You can check all the above points found using the graph of \[ y = - x^2 + 2 x + 3 \] shown below.

The points of intersection are solutions of the simultaneous equations \[ 2x + 3y = 7 \quad \text{and} \quad y = - 2 x^2 + 2 x + 5 \]. Substitute \( - 2 x^2 + 2 x + 5 \) for \(y \) in the equation \( 2x + 3y = 7 \) to obtain \[ 2x + 3(- 2x^2 + 2x + 5) = 7 \] Write the quadratic equation obtained above in standard form \[ -6x^2 + 8x + 8 = 0 \] Divide all the terms in the equation by 2. \[ -3x^2 + 4x + 4 = 0 \] Solution for x \[ x = 2 \quad , \quad x = -2/3 \] Substitute x for the previous solutions in \( 2x + 3y = 7 \) to find y. \[ x = 2 , y = 1 \quad \text{and} \quad x = -\frac{2}{3}, \quad y = \frac{25}{9} \] The intersection points are: \[ (2 , 1) \quad \text{and} \quad (-\frac{2}{3} , \frac{25}{9} ) \]. Check the answer graphically below.

The points of intersection of the two parabolas are solutions of the simultaneous equations \[ y = -(x - 3)^2 + 2 \quad and \quad y = x^2 - 4x + 1 \].
Substitute \( y \) by \( -(x - 3)^2 + 2 \) in the second equation: \[ -(x - 3)^2 + 2 = x^2 - 4x + 1 \] Expand, group like terms and write in standard form \[ -2x^2 + 10x - 8 = 0 \] Divide all terms by \( -2 \) \[ x^2 - 5x + 4 = 0 \] The solutions of the above quadratic equation are: \[ x = 1 \quad and \quad x = 4 \] Use one of the equations to find y:

\( x = 1 \) into the equation \(y = -(x - 3)^2 + 2 \) to get \( y = -(1 - 3)^2 + 2 = -2 \)

\( x = 4 \) into the equation \( y = -(x - 3)^2 + 2 \) to get \( y = -(4 - 3)^2 + 2 = 1 \)

Points of intersection are: \[ (1 , -2) \quad \text{and} \quad (4 , 1) \] Check the answer graphically below.

The points \((-1,-5)\) and \((2,10) \) are on the graph of the parabola \( y = 2 x^2 + b x + c\) and therefore are solutions to the equation of the parabola. Substituting by the coordinates of the two points, we obtain the equations:
\[ -5 = 2 (-1)^2 + b (-1) + c\] and \[10 = 2 (2)^2 + b (2) + c\] Rewrite the above system with unknowns \( b \) and \( c \) in standard form. \[ - b + c = - 7\]
\[ 2b + c = 2\] Solve the above system of equations to obtain: \( c = - 4 \) and \( b = 3\) Equation of the parabola that passes through the points \( (-1,-5)\) and \( (2,10)\) is given by: \[ y = 2 x^2 + b x + c = 2 x^2 + 3 x - 4\] Use a graph plotter to check your answer by graphing \( y = 2 x^2 + 3 x - 4 \) and Check that the graph passes through the points \( (-1,-5) \) and \((2,10)\).

The equation of a parabola with x-intercepts at \( x = 2 \) and \( x = -3 \) can be written as the product of two factors whose zeros are the x-intercepts as follows: \[ y = a(x - 2)(x + 3) \] We now use the y-intercept at (0, 5), which is a point through which the parabola passes, to write: \[ 5 = a(0 - 2)(0 + 3) \] Solve for \(a\) \[ a = - \dfrac{5}{6} \] Equation: \[ y = - \dfrac{5}{6} (x - 2)(x + 3)\] Graph \( y = - \dfrac{5}{6} (x - 2)(x + 3)\) and verify that the graph has x-intercepts at \( x = 2 , x = -3 \) and a y-intercept at \( y = 5\).

The points \( (0,3), (1,-4) \) and \( (-1,4) \) lie on the graph of the parabola \( y = a x^2 + b x + c \) and are therefore solutions to the equation of the parabola. Therefore, we write the system of 3 equations as follows: \[ (0,3) \Rightarrow c = 3 \quad (I) \] \[ (1,-4) \Rightarrow a + b + 3 = -4 \Rightarrow a + b = -7 \quad (II) \] \[ (-1,4) \Rightarrow a - b + 3 = 4 \Rightarrow a - b = 1 \quad (III) \] Equation (I) gives: \[ c = 3 \] Substitute 3 for c in equations (II) and (III) \[ a + b = -7 \] \[ a - b = 1 \] Solve the system for \( a \) and \( b \) to obatain \[a = - 3 \; , \; b = - 4 \] The equation of the parabola is given by: \[ y = a x^2 + b x + c = -3 x^2 - 4x + 3 \] Graph the graphs of \( y = -3 x^2 - 4x + 3 \) and verify that the graph passes through the points \( (0,3), (1,-4) \) and \( (-1 ,4) \).

The equation of the parabola, with vertical axis of symmetry, has the form \( y = a x^2 + b x + c \) or in vertex form \( y = a(x - h)^2 + k \) where the vertex is at the point \( (h , k)\) .

In this case it is tangent to a horizontal line \( y = 3 \) at \( x = -2 \) which means that its vertex is at the point \( (h , k) = (-2 , 3) \). Therefore, the equation of this parabola can be written as: \[ y = a(x - h)^2 + k = a(x - (-2))^2 + 3 = a(x + 2)^2 + 3 \] Its graph passes through the point \( (0 , 5) \) which must satisfy the equation: \[ 5 = a(0 + 2)^2 + 3 = 4 a+ 3 \] Solve the above for \( a \) \[ a = \dfrac{1}{2} \] Equation of the parabola is given by: \[ y = \dfrac{1}{2}(x + 2)^2 + 3 \] Sketch the graphs of \( y = \dfrac{1}{2} (x + 2)^2 + 3 \) and verify that the graph is tangent to the horizontal line \( y = 3 \) at \( x = -2 \) and also the graph passes through the point \( (0 , 5) \).

A line and a parabola are tangent if they have one point of intersection only , which is the point of tangency. The points of intersection are found by solving the system \( y = m x - 3 \quad and \quad y = 3 x^2 - x \) Substitute \( y \) by \( m x - 3 \) in the second equation: \[ mx - 3 = 3 x^2 - x \] Write as a standard quadratic equation: \[ 3 x^2 - x(1 + m) + 3 = 0 \] The discriminant of the above quadratic equation is given by: \[ \Delta = (1 + m)^2 - 4(3)(3) \] The line is tangent to the parabola if they have one point of intersection only, which means if: \[\Delta = 0 \] Hence the equation: \[(1 + m)^2 - 4(3)(3) = 0 \] solve for \( m \) \[(1 + m)^2 = 36 \] Solutions are: \[ m = 5 \quad and \quad m = -7 \] Use a graph plotter to check your answer by plotting the graphs of the lines: \( y = 5 x - 3 \) ( \( m = 5 \) solution ), \( y = -7 x - 3 \) ( \( m = 7 \) solution) and the parabola \( y = 3 x^2 - x\) and check that the two lines are tangent to the graph of the parabola \( y = 3 x^2 - x\).

The points of intersection are found by solving the system \[ y = 2 x + b \quad and \quad y = - x^2 - 2x + 1 \] Substitute \( y \) by \( 2 x + b \) in the second equation \[ 2x + b = - x^2 - 2x + 1 \] Write as a standard quadratic equation: \[ - x^2 - 4x + 1 - b = 0 \] The discriminant of the above equation is given by: \[ \Delta = (-4)^2 - 4(-1)(1 - b) = 20 - 4b \] The graphs of \( y = 2 x + b \) and \( y = - x^2 - 2 x + 1 \) have two points of intersection if \( \Delta \gt 0 \) (case of two real solutions of a quadratic equation), hence the inequality \[ 20 - 4 b \gt 0 \] Solve for \( b \) \[ b \lt 5 \] Use a graph plotter to check your answer by plotting graphs of \( y = - x^2 - 2 x + 1 \) and lines through equations \( y = 2 x + b \) for values of \( b \gt 5 \) , \( b \lt 5 \) and \( b = 5 \) to see how many points of intersection of the parabola and the line there are for each of these values of \( b \).

The points of intersection are found by solving the system \[ y = a x^2 + x \quad and \quad y = 3 x + 1 \] Substitute \( y \) by \( 3 x + 1 \) in the equation \( y = a x^2 + x \) \[ 3 x + 1 = a x^2 + x \] Write as a standard quadratic equation: \[ a x^2 - 2 x - 1 = 0 \] Discriminant of the quadratic equation: \[ \Delta = (-2)^2 - 4(a)(-1) = 4 + 4 a \] The graphs are tangent if they have a point of intersection (case for a solution of a quadratic equation) if \( \Delta = 0 \). Hence \[ 4 + 4 a = 0 \] Solve for \(a\) \[ a = -1 \] Parabola equation: \[ y = -x^2 + x \] Graph \( y = - x^2 + x \) and \( y = 3 x + 1 \) to verify the answer above.

Start with: \[ y = x^2 \] Shift 3 units to the left: \[ y = (x + 3)^2 \] Reflect about the x-axis: \[ y = -(x + 3)^2 \] Shift up 4 units: \[ y = -(x + 3)^2 + 4 \]

Given: \[ y = - x^2 + 4 x + 6 \] Rewrite in vertex form by completing the square: \[ y = - x^2 + 4 x + 6 = - (x - 2)^2 + 10\] Beginning: \[ y = x^2\] Shift 2 units to the right: \[ y = (x - 2)^2\] Reflect about the x-axis: \[ y = -(x - 2)^2 \] Shift up 10 units: \[ y = -(x - 2)^2 + 10 \]

Any point identified on the given graph can be used to find the equation of the parabola. However, using the \( x \) and \( y \) intercepts and the vertex are better ways to find the equation of the parabola whose graph is shown below.

Two methods are presented to solve the problem:

method 1:
The graph has two x-intercepts: (-5, 0) and (-1, 0)
Use the two x-intercepts at (-5, 0) and (-1, 0) to write the equation of the parabola as a product of two linea factors: \[ y = a(x + 1)(x + 5)\] Use the y-intercept at (0, -5) to write \[ - 5 = a(0 + 1)(0 + 5) = 5 a\] Solve for \(a \) \[ a = -1 \] Write the equation of the parabola: \[ y = -(x + 1)(x + 5) = - x^2 - 6 x - 5 \]

method 2:

Use the vertex at \( ( h , k) = (-3 , 4) \) to write the equation of the parabola in vertex form as follows: \[ y = a(x - h)^2 + k = a(x + 3)^2 + 4 \] Use the y-intercept (0, -5) to find \(a\). \[ - 5 = a(0 + 3)^2 + 4 \] Solve the above for \(a\): \[ a = -1 \] The equation of the parabola is given by \[ y = -(x + 3)^2 + 4 = - x^2 -6 x - 5 \]