QUESTION IMAGE
Question
graph the function $f(x) = x^2 + 3$. then use the line tool to draw the line $x = 0$ and then use the dot tool to plot the point $(0, f(0))$.
Step1: Analyze the function type
The function \( f(x) = x^2 + 3 \) is a quadratic function (a parabola). The standard form of a quadratic function is \( y = ax^2 + bx + c \), here \( a = 1 \), \( b = 0 \), \( c = 3 \). Since \( a>0 \), the parabola opens upwards. The vertex of the parabola \( y = ax^2+bx + c \) is at \( x=-\frac{b}{2a} \), substituting \( a = 1 \), \( b = 0 \), we get \( x = 0 \). Then \( f(0)=0^2 + 3=3 \), so the vertex is at \( (0, 3) \).
Step2: Find some key points
To graph the parabola, we can find a few more points. For \( x = 1 \), \( f(1)=1^2 + 3=4 \), so the point is \( (1, 4) \). For \( x=- 1 \), \( f(-1)=(-1)^2+3 = 4 \), so the point is \( (-1, 4) \). For \( x = 2 \), \( f(2)=2^2+3 = 7 \), point \( (2, 7) \). For \( x=-2 \), \( f(-2)=(-2)^2 + 3=7 \), point \( (-2, 7) \).
Step3: Draw the parabola
Plot the vertex \( (0, 3) \), and the other points \( (1, 4) \), \( (-1, 4) \), \( (2, 7) \), \( (-2, 7) \) etc., then draw a smooth curve (the parabola) through these points.
Step4: Draw the line \( x = 0 \)
The line \( x = 0 \) is the y - axis. We can use the line tool to draw a vertical line along the y - axis (passing through all points with \( x = 0 \)).
Step5: Plot the point \( (0,f(0)) \)
We already found that \( f(0)=3 \), so the point is \( (0, 3) \). We can use the dot tool to plot this point (which is also the vertex of the parabola).
(Note: Since this is a graphing problem, the final answer is the graphical representation as described above. The key elements are the parabola \( y=x^2 + 3 \), the line \( x = 0 \) (y - axis), and the point \( (0, 3) \) plotted on the graph.)
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The graph of \( f(x)=x^2 + 3 \) is a parabola opening upwards with vertex at \( (0, 3) \), the line \( x = 0 \) is the y - axis, and the point \( (0, 3) \) is plotted. (The actual graph should be drawn as per the steps above on the given coordinate grid.)