Question Video: Identifying the Graph of a Quadratic Function Mathematics

Choose the graph that represents the function π(π₯) = 2π₯Β² + 2. [A] Graph A [B] Graph B [C] Graph C [D] Graph D [E] Graph E

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Video Transcript

Choose the graph that represents the function π of π₯ equals two π₯ squared plus two.

The function weβve been given is a quadratic function of the form π of π₯ equals ππ₯ squared plus π. In this case, the values of π and π are each equal to two. From this, we can deduce that the graph that represents this function is a parabola with the π¦-axis as its a line of symmetry. Because the value of π, the coefficient of π₯ squared, is positive, the parabola will be U shaped. It will open upwards. We can find the π¦-intercept of the curve by evaluating π of zero because π₯ is equal to zero on the π¦-axis. This gives π of zero equals two multiplied by zero squared plus two. Thatβs zero plus two, which is equal to two.

We can also recall that in general the π¦-intercept of the function π of π₯ equals ππ₯ squared plus π is π. So our value of two is consistent with the value of two as the constant term in our given function.

So now, we know that weβre looking for a U-shaped parabola with a π¦-intercept of two. As this point is on the π¦-axis, which is the line of symmetry for this parabola, the point with coordinates zero, two will also be the vertex of the curve. Based on this, we can rule out options (B), (D), and (E). Option (B) is a U-shaped parabola with the π¦-axis as its line of symmetry, but its vertex is the point zero, zero. (D) and (E) are n-shaped parabolas or parabolas that open downwards. So they correspond to quadratic functions with a negative coefficient of π₯ squared.

Weβre left with options (A) and (C). These are both parabolas which open upwards, they both have the π¦-axis as their line of symmetry, and they each have their vertex at the point zero, two. To decide which is the correct graph, we can choose any other point that lies on the curve and test whether the coordinates of this point satisfy the function π of π₯ equals two π₯ squared plus two.

For graph (A), we can use the point with coordinates one, three. Evaluating π of one, we have two multiplied by one squared plus two. Thatβs two plus two, which is equal to four. And as this isnβt equal to three, this tells us that graph (A) is not the correct graph to represent the function π of π₯ equals two π₯ squared plus two. If we look at graph (C), however, we can see that the point with coordinates one, four does lie on this curve.

We can perform a further check by using another point, perhaps the point with coordinates two, 10. Evaluating π of two, we have two multiplied by two squared plus two. Two squared is four. Multiplying by two gives eight, and adding two gives 10. So this confirms that the point with coordinates two, 10 satisfies the function π of π₯ equals two π₯ squared plus two.

So we found that the graph that represents the function π of π₯ equals two π₯ squared plus two is graph (C). It has the correct shape, the correct line of symmetry, the correct vertex, and weβve checked that the coordinates of two other points that lie on the curve satisfy the given function.