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Find a quadratic function given its Graph. Examples with detailed solutions are presented. A tutorial with examples on graph of quadratic functions might help in understanding the present examples on finding quadratic equations.
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The general form of a quadratic function is written as:
\[ f(x) = ax^2 + bx + c \]A quadratic function \( f \) in vertex (or standard) form is written as:
\[ f(x) = a(x - h)^2 + k \]where \( h \) and \( k \) are the \( x \)- and \( y \)-coordinates, respectively, of the vertex (minimum or maximum) point of the graph.
The graph of \( f \) is a parabola with the vertical line \( x = h \) as its axis of symmetry.
The relationship between \( h \) and \( k \) is given by:
\[ h = \frac{-b}{2a} \quad \text{and} \quad k = f(h) = c - \frac{b^2}{4a} \]Find the quadratic function \( f \) whose graph is shown below.
Let \( h \) and \( k \) be the coordinates of the vertex of the graph of the function \( f \).
From the graph, the vertex (minimum point) is identified as \( (h, k) = (0, 2) \). Hence, the vertex form of the function \( f \) can be written as:
\[ f(x) = a(x - h)^2 + k = a(x - 0)^2 + 2 = ax^2 + 2 \]The point \( (1, 3) \) lies on the graph of \( f \), and it can be used to find the coefficient \( a \).
\[ f(1) = a(1)^2 + 2 = 3 \]Solving for \( a \):
\[ a + 2 = 3 \Rightarrow a = 1 \]Therefore, the function \( f \) is:
\[ f(x) = x^2 + 2 \]Find the quadratic function \( g \) whose graph is shown below and evaluate \( g(-3) \).
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The vertex of the graph of function \( g \) is a maximum point located at \( (h, k) = (0, -1) \). Hence, the function \( g \) in vertex form is written as:
\[ g(x) = a(x - h)^2 + k = a(x - 0)^2 - 1 = ax^2 - 1 \]The coefficient \( a \) will now be found using the point \( (1, -2) \), which lies on the graph of \( g \).
\[ g(1) = a(1)^2 - 1 = -2 \]Solve the equation above for \( a \):
\[ a - 1 = -2 \quad \Rightarrow \quad a = -1 \]Hence, the function \( g(x) \) is given by:
\[ g(x) = -x^2 - 1 \]Evaluate \( g(-3) \):
\[ g(-3) = -(-3)^2 - 1 = -9 - 1 = -10 \]Find the quadratic function\( l \) whose graph is shown below and calculate the x-intercepts of the graph.
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The graph of function \( l \) has a vertex (maximum point) located at \( (h, k) = (2, 1) \). Function \( l \) in vertex form is written as:
\[ l(x) = a(x - h)^2 + k = a(x - 2)^2 + 1 \]We use the y-intercept \( (0, -7) \) of the graph of \( l \) to find the coefficient \( a \) as follows:
\[ l(0) = a(0 - 2)^2 + 1 = -7 \]Solve the equation for \( a \):
\[ a \cdot 4 + 1 = -7 \quad \Rightarrow \quad 4a = -8 \quad \Rightarrow \quad a = -2 \]Function \( l(x) \) is therefore given by:
\[ l(x) = -2(x - 2)^2 + 1 \]We now calculate the x-intercepts by solving the equation:
\[ -2(x - 2)^2 + 1 = 0 \]Move 1 to the other side and divide by -2:
\[ (x - 2)^2 = \frac{1}{2} \]Extract the square root on both sides:
\[ x - 2 = \pm \sqrt{\frac{1}{2}} \] \[ x = 2 \pm \sqrt{\frac{1}{2}} \]Therefore, the x-intercepts of the function are located at the points:
\[ \left(2 - \sqrt{\frac{1}{2}}, 0\right) \quad \text{and} \quad \left(2 + \sqrt{\frac{1}{2}}, 0\right) \]Find the quadratic function s in standard form whose graph is shown below.
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The graph of the function \( s \) has two x-intercepts: \( (-1, 0) \) and \( (2, 0) \). This means that the equation \( s(x) = 0 \) has two solutions: \( x = -1 \) and \( x = 2 \).
Hence, the function \( s(x) \) can be written as the product of two factors:
\[ s(x) = a(x + 1)(x - 2) \]We now use the y-intercept \( (0, -4) \) of the graph of \( s \) to determine the coefficient \( a \):
\[ s(0) = a(0 + 1)(0 - 2) = -4 \]Solve the above equation for \( a \):
\[ a = 2 \]The function \( s(x) \) is given by:
\[ s(x) = 2(x + 1)(x - 2) \]Now, expand and simplify to write \( s(x) \) in standard form:
\[ s(x) = 2(x^2 - x - 2) = 2x^2 - 2x - 4 \]Find the quadratic function m in standard form whose graph is a parabola with an axis of symmetry given by the vertical line x = -3 as shown below.
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The graph has an axis of symmetry given by the vertical line \( x = -3 \). Hence, the x-coordinate \( h \) of the vertex is equal to -3, and the function \( m(x) \) may be written as:
\[ m(x) = a(x + 3)^2 + k \]We now have two unknowns, \( a \) and \( k \), to determine. Using the points \( (-5, 0) \) and \( (-2, -\frac{3}{2}) \) shown on the graph of \( m \), we can write two equations:
Simplifying both equations, we obtain the following system of linear equations:
\[ 4a + k = 0 \] \[ a + k = -\frac{3}{2} \]Solving this system, we find:
\[ a = \frac{1}{2}, \quad k = -2 \]Substitute these values into the vertex form of the function:
\[ m(x) = \frac{1}{2}(x + 3)^2 - 2 \]Now expand and rewrite \( m(x) \) in standard form:
\[ m(x) = \frac{1}{2}x^2 + 3x + \frac{5}{2} \]Find the quadratic function w in standard form whose graph is a parabola shown below.
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The quadratic function \( w(x) \) in standard form is written as follows:
\[ w(x) = ax^2 + bx + c \]We need to find the coefficients \( a \), \( b \), and \( c \). We use three points on the graph of \( w \) to write a system of three equations:
The point \( (0, -\frac{1}{6}) \) gives:
\[ w(0) = a(0)^2 + b(0) + c = -\frac{1}{6} \quad \text{(eq 1)} \]The point \( (1, 0) \) gives:
\[ w(1) = a(1)^2 + b(1) + c = 0 \quad \text{(eq 2)} \]The point \( (3, \frac{10}{3}) \) gives:
\[ w(3) = a(3)^2 + b(3) + c = \frac{10}{3} \quad \text{(eq 3)} \]Equation 1 simplifies to:
\[ c = -\frac{1}{6} \]Substitute \( c = -\frac{1}{6} \) into equations 2 and 3:
\[ a + b = \frac{1}{6} \] \[ 9a + 3b = \frac{7}{2} \]Solving the above system of equations yields:
\[ a = \frac{1}{2}, \quad b = -\frac{1}{3} \]Substituting \( a \), \( b \), and \( c \) into the standard form of the quadratic function gives:
\[ w(x) = \frac{1}{2}x^2 - \frac{1}{3}x - \frac{1}{6} \]